These lectures have been designed for a course given at the Ecole Normale of Parisin 2017.
I used the following books to prepare these lectures
1. Semi-Riemannian Geometry by O’Neil [O’N83],
2. Global Lorentzian Geometry by Beem, Ehrlich and Easley [O’N83],
3. The Cauchy problem in General Relativity by H. Ringström [Rin09],
4. Mathematical problems of General Relativity I. by D. Christodoulou [Chr08].
Most of the beginning follows [O’N83]. The introduction to the Einstein equationsis more personal. I used [O’N83] in Section 10. The construction of the time andtemporal functions is taken from [Rin09], which itself followed the papers [BS05],[O’N83], [BS03]. We claim no originality in these notes, neither in their content, norin the way it is written (but we do hope the students will find these notes useful! ).
For a student who would be interested in an introduction to General Relativityfrom a Physics point of view, a good reference is D’Inverno, Introducing Einstein’sRelativity [d1992introducing].
The goals of the lectures are to
1. define the basic objects of Riemannian and Lorentzian geometry,
2. give some of the fundamental results in Riemannian and Lorentzian geome-try.
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3. introduce the student to some analysis (pdes) problems and to the (by now)standard techniques to tackle them. More specifically, we will focus on hyper-bolic pdes arising naturally in the context of Lorenzian geometry.
To set up a target, one of the main goal of these lectures will be to prove thefollowing theorem1
Theorem 1.1. Let (M , g ) be a smooth, oriented, time-oriented Lorentzian manifold.Assume that (M , g ) is globally hyperbolic and that Σ is a smooth Cauchy hypersurfacewith future unit normal n. Let ψ0, ψ1 be smooth functions on Σ and F a smoothfunction on M. Then the Cauchy problem
ägψ = F,
ψΣ = ψ0,
n(ψΣ) = ψ1
admits a unique smooth solution ψ.
Obviously, we need to define everything above. In particular, äg will be a wavetype operator generalizing the classical wave operator −∂2
t +∆ and Σwill be replace-ment for the hypersurface t = 0 in Rn+1. The pure pde aspect of the problem willactually be easy (say easier than solving a general hyperbolic pde in Rn+1 as seen inthe PDE class). The hard part will be essentially to split the manifold on small pieces,on which we can solve the wave equation, and then glue everything together. To al-low such a cut and paste argument we need the geometric assumption of "globallyhyperbolicy" and a large part of the lectures will try to prepare us to define and usethat assumption.
2 Review of differential geometry
I expect the students to already be aquainted with most of the content of this section.Most proofs are left to the reader.
The basic object we will need is that of a differential manifold.
2.1 Differential manifolds
Definition 2.1. Let M be a topological space. A coordinate system on M is an home-omorphism ξ from an open set U ⊂ M into an open set of Rn .
Two coordinate ξ1 and ξ2 systems overlap smoothly if ξ1 ξ−12 and ξ2 ξ−1
1 aresmooth (exercise: on which domains of definition ? ).
Definition 2.2. An atlas on M is a collection of coordinate systems such that the unionof all their domains of definition cover M and such that any two coordinate systemsoverlap smoothly.
Exercise 2.1. Recall the definition of the dimension of M. What if we replace smoothby continuous ? ( a lot harder )
1This theorem could be attributed to Leray, following his 1952 lectures on hyperbolic pdes. The notionof global hyperbolicity used in his lectures is slightly different and in fact he proves much more, since hismain interest was general hyperbolic pdes, not just the wave equation.
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An atlas A is called complete if it contains each coordinate system that overlapssmoothly with every coordinate systems of A.
Lemma 2.1. Every atlas is contained in a unique complete atlas.
Definition 2.3. A smooth manifold is a Hausdorff space endowed wih a completeatlas.
We will also assume in all of these lectures that all manifolds considered are sec-ound countable (i.e. admits a countable basis of open sets).
Exercise 2.2. 1. what does Hausdorff mean again ?
2. what would be a good definition of an open submanifold of M ?
3. Same with product manifolds of M and N ?
2.2 Differential mappings
Definition 2.4. Let M and N be manifolds. A mappingφ : M → N is smooth if for anycoordinate systems (UM ,ξM ) and (UN ,ξN ) of M and N , we have ξN φξ−1
M smooth.
Notation : We will write F (M) for the set of smooth real valued functions of M .
2.3 Tangent vectors
Let p ∈ M (M manifold. ) A tangent vector X at p is a real-valued function
X : F (M) →R,
such that
1. X is R-linear.
2. X ( f g ) = f (p)X (g )+ g (p)X ( f ).
(Think of X as a derivation at p).Let Tp M denote the set of all tangent vectors at p. Tp M is a real vector space (for
which operations ?) called the tangent space to M at p.
Definition 2.5. Let (U ,ξ = (xα)) be a coordinate system of M and p ∈ U . Let f ∈F (M). Then,
∂xi f (p) := ∂ f ξ−1
∂xi(ξ(p)).
Let[∂xi
]|p : f → ∂xi f (p). One then easily check that
[∂xi
]|p ∈ Tp M . Often, we
will drop the p index and simply write ∂xi .
Lemma 2.2. Let X ∈ Tp M , f , g ∈F (M). Then,
1. if f = g on some neighborhood of p, then X ( f ) = X (g ).
2. if f = const on some neighborhood of p, then X ( f ) = 0.
This lemma implies that tangent vectors are purely local objects.
Theorem 2.1. If ξ= (xα) is a coordinate system of M, then (∂xα ) form a basis of Tp M.In particular, dimTp M = dim M.
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2.4 The differential of a map
Definition 2.6. Let M , N be smooth manifolds and φ : M → N smooth. For any X ∈Tp M, define YX : F (N ) → R as Y (g ) = X (g φ). Then, Y ∈ Tφ(p)M. The map sendingeach such X to YX is called the differential (or pushforward) of φ at p and denoteddφp . (dφp : Tp (M) → Tφ(p)N ).
Exercise 2.3. Show that dφp is linear and compute it in local coordinates.
A specific case is that of N =R and f ∈F (M). By definition,
d fp : Tp M → T f (p)R'R
so it can be considered as an element of (Tp M)?.
2.5 The inverse function theorem
Theorem 2.2. Let φ : M → N be a smooth mapping. The differential map dφp atsome p ∈ Tp M is a linear isomorphism if and only if there is a neighborhood N ofp in M such that φ|N is a diffeomorphism from N onto a neighborhood φ(N ) ofφ(p) ∈ N .
Proof. This is a consequence of the usual inverse function theorem for subsets ofRn .
2.6 Curves
A curve is smooth map γ : I → M , where I is an non-empty interval. Given a curveγ, we write
γ(s) = dγ
d s(s) = dγs [
d
d s]|s ∈ Tγ(s)M .
A piecewise smooth curve is a continuous map γ : I → M , where I is an non-empty interval such that if I = [a,b], then there exists a finite sequence a = t0 < t1.. <tk = b such that each curve segment γ|[ti ,ti+1] is a smooth curve, while for an intervalwith open endpoints, say I = [a,b), we require that for each non-empty closed inter-val J ⊂ I , the restriction γ|J is piecewise smooth (this implies that the break pointshave no cluster point in I ).
2.7 Vector fields and tangent bundle
Let T M = ⊔p∈M Tp M be the disjoint union of all the tangent planes of M . We call
T M the tangent bundle of M . Associated to T M is the a canonical projection de-fined by
π : T M → M
(p, X ) → p
T M can be given a natural manifold structure as follows. Let (xα) be a coordinatesystem of M . Let p ∈ Tp M . For any v ∈ Tp M , we have
v = vα∂xα
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The (vα) then forms a global coordinate system on Tp M (called conjugate to the(xα) ). Given (p, X ) ∈ T M , let ξ(p, X ) = (ξ(p), Xα), then ξ defines a local coordinatesystem of T M and one easily verifies that any two such local coordinate systemsoverlap smoothly.
The tangent bundle is fact the basic example of a vector bundle over M .A vector field on M is by definition a section of T M , i.e. a map of the form
X : M → T M
satisfying π X = I dM . We will write Xp to denote the second component of X (p),i.e. Xp ∈ Tp M , ∀p ∈ M .
Given a vector field X and f ∈ F (M), the function X ( f ) is by definition X ( f ) :p → Xp ( f ). The set Γ(M) of vector fields on M can be naturally given the structureof a module over the ring F (M).
A vector field X is smooth if X ( f ) is smooth for any f .Given a local coordinate system (xα), we associate the vector fields ∂xα , defined
so that [∂xα ]p ( f ) = [∂xα ]|p ( f ).For any vector field X , we have
X =∑α
X (xα)∂xα .
Identification with derivation
A derivation on F (M) is a R-linear function D : F (M) → F (M) which satisfies theusual Leibniz property
D( f g ) =D( f ).g + f .D(g ).
Exercise 2.4. Prove that every vector field defines a derivation and that every deriva-tion can be realized as a vector field.
The Lie bracket
Definition 2.7. Let X ,Y be vector fields. Then, we define the vector field [X ,Y ] by
[X ,Y ]p ( f ) = Xp (Y ( f ))−Yp (X ( f )).
Push forward of a vector field by diffeomorphism
Definition 2.8. Let φ be a diffeomorphism φ : M → N and X a vector field on M.Then, we can define a vector field on N by the formula
(dφX )g = X (g φ)φ−1, ∀g ∈F (N ).
Note that this definition breaks down if φ is merely is smooth function.
2.8 One forms
Let Tp M∗ := (Tp M)∗ be the vector space dual of Tp M . Its element are called cov-ectors. Let T M∗ denote the cotangent bundle, obtained as above by the disjointunion of the Tp M∗. A one form is then a smooth section of T M∗ (ex: give T M∗ thestructure of a smooth vector bundle). The set of 1-forms is denotedΛ1(M).
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Definition 2.9. Let f ∈ F (M). Then, we define d f ∈ Λ1(M) as d f (X ) = X ( f ) for alltangent vector X to M.
Given a coordinate system (xα) on M , for each α, the differential of the functionxα, d xα is thus a 1-form.
Let ∂xα denotes the coordinate vector fields, then the (d xα) provides at eachpoint a dual basis to the (∂xα ). Moreover, we have
d f =∑α∂xα ( f )d xα.
2.9 Einstein summation convention2
By convention, we will remove the∑
symbols everytime we have repeated indices.For instance, in the previous formula, we write
d f = ∂xα ( f )d xα.
Again by convention, indices corresponding to the components of vectors are low-ered, and indices corresponding to components of covectors are put in higher po-sitions. With this convention, a repeating sum will always involve indices up anddown (and no sum with indices only up should appear).
2.10 Submanifolds
Definition 2.10. A manifold P is a submanifold of M if
• P ⊂ M and P has the induced topology of M.
• The inclusion map i : P → M is smooth and its differential is injective.
Exercise 2.5. Consider M = R2 and P = R(0,1)+R(1,0) as a topological subspace ofM.
1. We consider the mapψ : P →R
defines by ψ(y,0) = y, ψ(0, x) =−x. Prove that ψ is homeomorphism and thusthat it defines a global chart on P. It follows that P is manifold.
2. Prove that the inclusion map iP : P → M is not smooth.
3. Let now Q = (t ,0) : t ∈ [0,1)∪ (0, t ) : t ∈ [0,1)). Again, we consider Q as atopological subspace of R2. We us define the map
φ : Q →R
by φ(e−1/t 2
,0)= t , φ
(0,e−1/t 2
)= −t , φ(0,0) = 0. We recall that t 6= 0 → e−1/t 2
can be extended smoothly by 0 at t = 0. Prove that φ is homeomorphism andthus defines a global chart on Q, making Q a manifold.
4. Prove that the inclusion map iQ : Q → M is smooth but that its differential isnot injective.
2According to the usal folklore knowledge, Einstein thought of this as his greatest invention. Let meknow if you know an exact citation on this.
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5. The restriction of ψ to Q defines another potential candidate for a coordinatesystem on Q. Prove that the two coordinates systems do not overlap smoothly.
Proposition 2.1. If P m is a submanifold of M n , then for all p ∈ P, there exists a coor-dinate system (x1, .., xm , xm+1, .., xn) defined on U ⊂ M such that p ∈U and
P = (xm+1 = 0, .., xn = 0).
Proof. Use the inverse function theorem.
We call such a coordinate system adapted to P .
Proposition 2.2. A subset P ⊂ M is a submanifold of M if for each point p of P, thereis a coordinate system of M adapted to P.
Let P be a submanifold of M , p ∈ P and denote the inclusion map by i . Then
dip : Tp P → Tp M .
Since dip is injective, we will identify Tp P with dip (Tp P ) and therefore view Tp P asa vector subspace of Tp M . A vector field X on M is then tangent to P , if for all p ∈ P ,Xp ∈ Tp P .
2.11 Immersions, submersions, embedding
Definition 2.11. An immersion is a smooth mapφ : M → N such that dφp is injectivefor all p ∈ M.
Definition 2.12. An embedding of P into M is a immersion φ : P → M such that
1. φ is injective.
2. the induced map φ|φ(P ) :φ→φ(P ) is an homeomorphism.
Clearly, the restriction of any immersion to any sufficiently small open set is animbedding.
Exercise 2.6. 1. If P is a submanifold of M then the inclusion map i : P → M isautomatically an imbedding.
2. Consersely, letφ : P → M be an imbedding. Giveφ(P ) a manifold structure suchthat φ : P →φ(P ) is a diffeomorphism. Then φ(P ) ⊂ M and the inclusion is animmersion, so that φ(P ) is a submanifold of M.
Definition 2.13. Let ψ : M m → N n be a smooth map and let q ∈ N . q is said to beregular provided that dψp is onto, for every p ∈ψ−1(q).
Lemma 2.3. If q ∈ψ(M) is regular, then ψ−1(q) is a submanifold of M and dim M =dim N +dimψ−1(q).
By definition, a hypersurface is a submanifold of codimension 1. They can beconstructed from the previous lemma with N =R.
Corollary 2.1. Let f : M → R and c ∈ Im( f ). If d fp 6= 0 for every p ∈ f −1(c), thenf −1(c) is a hypersurface of M (called a level set of f ).
Definition 2.14. A submersionψ : M → B is a smooth mapping onto B such that dψp
is onto for all p ∈ M.
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2.12 Integral curves
A curve α is an integral curve of X ∈ Γ(M) if α= Xα.In coordinates, an integral curve is a solution to
d(xi α)
d s= X i α.
By Cauchy-Lipshitz, for any X ∈ Γ(M) and any p ∈ M , there exists a unique maximalintegral curve of X starting at p.
Exercise 2.7. Prove that for any vector field X such that Xp 6= 0, there exists alocal coordinate system (xα) at p such that X = ∂x1 .
Definition 2.15. Let p ∈ M. A local flow of X is a neighborhood U of p, a neighbord-hood I of 0 ∈R and a map φ of the form
φ : I ×U → M
(t , p) → φt (p)
where φt (p) is the value at time t of the solution of the integral curve equation for Xwith initial data (0, p).
In fact, still by standard ode theory, there exists an open set D ⊂R×U containing0×U , such that φ is well-defined on D .
A vector field is called complete if its flow is defined on R×M , or another words,if its integral curves are all defined for all times.
2.13 Tensor fields
Definition 2.16. A tensor field T of type (r, s) over a manifold is a map of the form
T : (Λ1(M))r × (Γ(M))s →F (M)
which is F (M)-multilinear.
We will denote by T rs the set of tensor fields of type (r, s).
Definition 2.17 (Tensor product). Given T1 a tensor field of type (r1, s1) and T2 oftype (r2, s2), we define T1 ⊗T2 as
Exercise 2.8. • Define a (r, s) tensor bundle and tensor fields as a section of thatbundle.
• Recall that the Lie Bracket [., .] maps a pair of vector fields to a vector field. Thus,we can define a map ψ on (Λ1(M))1 × (Γ(M))2 by
ψ(ω, X ,Y ) =ω ([X ,Y ]) .
Prove that ψ does not define a tensor (1,2) tensor field.
• Show that (0,0) tensor fields, (1,0) tensor fields and (0,1) tensor fields can beidentified with functions, vector fields and one-forms respectively.
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2.14 Tensors at a point
First a classical lemma.
Lemma 2.4. Given any neighborhood U of a point p ∈ M, there is a function f ∈F (M), called a bump function at p, such that
1. 0 ≤ f ≤ 1 on M,
2. f = 1 on some neighborhood of p,
3. The support of f is included in U
supp f := p ∈ M : f (p) 6= 0 ⊂U .
Proof. Exercise.
Proposition 2.3. Let p ∈ M and A ∈ T rs . Let (θi )1≤i≤s , (νi )1≤i≤s be 2s one forms such
that θip = νi
p for all i . Similarly, let (Xi ), (Yi ) for 1 ≤ i ≤ r be 2r vector fields on M suchthat Xi (p) = Yi (p).
Then,A(θ1,θ2, ..,θs , X 1, .., X r )(p) = A(ν1, ..,νs ,Y 1, ..,Y r )(p).
Proof. We do the proof in the case of tensor fields of type (1,1). Let θ be a one formand X be a vector field. Let (U , xα) be a coordinate system around p and let f be abump function at p with support in U . We can decompose X and θ,
X = Xα∂xα , (1)
θ = θαd xα, (2)
Here any of the components Xα, θα are smooth functions on U . Moreover, f Xα,f θα, f ∂xα , f d xα are all smooth in U and can be extended trivially to the whole of Mby 0. (We denote the resulting object by the same letters).
Thus, by F (M)-multilinearity
A( f 2θ, f 2X ) = f θα. f X βA( f d xα, f ∂xβ ).
Evaluated at p, this gives
f 4(p)A(θ, X )(p) = A( f 2θ, f 2X )(p) = f (p)θα(p) f (p)X β(p)A( f d xα, f ∂xβ )(p),
so that
A(θ, X )(p) = θα(p)X β(p)A( f d xα, f ∂xβ )(p).
This implies that given a tensor field A of type (r, s), we can define a map
Ap : (T?p M)r × (Tp M)s →R
as follows. Given ω1, ..,ωr ∈ Tp M? and x1, .., xs ∈ Tp M , let θi , X i be respectively oneforms and vector fields such that θi (p) =ωi , Xi (p) = xi . Then, define
Ap (ω1, ..,ωr , x1, .., xs ) = A(θ, X )(p).
At each p, Ap is R-multilinear (it is thus a tensor over the vector space Tp M).Moreover, the above proposition implies that given any open set U of M , the
restriction A|U of A to U is a well defined tensor field on U .In particular,
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Definition 2.18. Let (U ,ξ= (xα)) be a coordinate system. Then, for any tensor A thecomponents of A (relative to ξ) are the real-valued functions
Ai1..irj1... js
= A(d xi1 , ..,d xir ,∂ j1 , ...,∂ js ).
Lemma 2.5. With the above notation, the tensor A verifies on U
A = Ai1..irj1.. js
∂i1 ⊗ ..⊗∂ir ⊗d x j1 ..⊗d x j s .
Proof. Show that the left and right-hand side take the same value on each (d xi1 , ..d xir ,∂ j1 ..∂ js )and use the F (U ) multi-linearity.
Tensor transformation rules
We give the transformation rule for tensor of type (1,1) and leave the general case asan exercise.
Lemma 2.6. Let A be a tensor field of type (1,1) and let (xα), (y i ) be two coordinatesystems defined on a common open set U . Denote the components of A with respectto (x) as aα
βand with respect to (y) as bi
j , then
aαβ = ∂xα
∂y i
∂y j
∂xβbi
j .
Remark 2.1. In the above formula, the interpretation of say ∂xα
∂y i is that for each α, xα
is a function on U and for each i , ∂y i is a vector field. Thus, ∂y i (xα)(= ∂xα
∂y i ) is well
defined.
Proof. Just use the previous formula giving A in terms of its components and applythe change of coordinate rules to each of ∂xα , d xα.
Remark 2.2. Let (U ,ξ= (xα)) be a coordinate system and let Xα(x1, .., xn) be smoothfunctions defined on ξ(U ) ⊂ Rn . Then X := Xα ξ.∂xα defines a vector field on U .Now let ξ′ = (yα) be another coordinate system on U and let (X ′)α be another set offunctions defined on ξ′(U ). Then X ′ := (X ′)α ξ′.∂yα = X if and only if the usual rulefor changes of coordinates applies to Xα and X ′α. Thus, if for any coordinate systemξ on U , we can construct components Xα (defined on ξ(U )), we can then define aunique tensor field provided that the constructed components satisfy the usual rulefor coordinate transformations.
Remark 2.3. The components of tensor need not be only with respect to coordinateinduced basis. More precisely, a basis of (local) vector fields is a set of (local) vectorfields Xi such that at each p, the Xi (p) form a basis of Tp M. One can similarly intro-duce a basis of one-forms (which may or may not correspond to the dual basis of thebasis of vector fields), and finally decompose all tensors with respect to these basis.
2.15 Contraction
The contraction is an operation on tensor fields taking (r,s) types to (r-1,s-1) types.It can be viewed as a trace operation. More precisely,
Lemma 2.7. There is a unique F (M)-linear function C defined on tensor of type (1,1)such that C (X ⊗θ) = θ(X ), ∀(X ,θ) ∈ Γ(M)×Λ(M).
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Proof. In coordinates, given A βα the components of a tensor of type (1,1), check
that C (A) = A αα necessarily and that this defines C properly.
Exercise 2.9. Define a contraction C ji over the indices i and j for tensors of type (r, s)
and 1 ≤ i ≤ r , 1 ≤ j ≤ s.
2.16 Pull-back
Let φ : M → N be a smooth map. Let ω ∈Λ1(N ). Then, we can define a one-form onM by
∀p ∈ M , v ∈ Tp M , φ?(ω)(v) =ωφ(p)(dφp v).
Similarly, if A is a (0, s) tensor field, then define for each p on M
φ?(A)(v1, .., vs ) = Aφ(p)(dφp v1, ..,dφp vs ).
2.17 Vector bundles
The tangent bundle can be viewed as an example of a vector bundle, defined as fol-lows.
Definition 2.19. A k-vector bundle (E ,π) over a manifold M consists in a manifold Etogether with a smooth map
π : E → M ,
such that
1. for all p ∈ M, each fiber π−1(p) is a k-dimensional vector space.
2. for each p ∈ M, there is a neighborhood U of p and a diffeomorphism
φ : U ×Rk →π−1(U ),
such that for each q ∈U , the map v →φ(q, v) is a linear isomorphism from Rk
to π−1(q). (In particular πφ(q, v) = q.)
If M is a manifold, p ∈ M and ξ = (U , xα) is a local coordinate system near p,then for any point (q, vq ) ∈ T M , we have the decomposition vq = V α(q)[∂xα ]q . Wecan then consider
φ : U ×Rn → π−1(U )
(q,V ) → (q,V α[∂xα ]q )
which shows that T M is an example of a vector bundle.A section of a vector bundle E is a map X : M → E such that πX = I dM .
3 Semi-Riemannian manifolds
3.1 Semi-Riemannian metrics
Definition 3.1. A scalar product g on a finite dimensional real vector space V is anon-degenerate symmetric bilinear form on V .
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Non-degenerate means that, for each v 6= 0 ∈ V , ∃x ∈ V , such that g (v, x) 6= 0.Equivalently, any matrix representing g is invertible.
We say that v and w are orthogonal, denoted v ⊥W , if g (v, w) = 0.If W is a subspace of V , we define as usual W ⊥ and we have
Lemma 3.1.
dimW +dimW ⊥ = dimV , (3)
(W ⊥)⊥ = W. (4)
Exercise 3.1. Prove the lemma. Does (4) still hold for infinite dimensional vectorspace V equipped with non-degenerate symmetric bilinear form ?
Because g (v, v) can be negative, the norm |v | of a vector is defined to be
|v | = |g (v, v)|1/2,
(but it does not define a norm in the sense of normed vector space unless g is posi-tive definite). A unit vector is a vector of norm 1, i.e. such that g (v, v) =±1. As usuala set of mutually orthogonal unit vectors is said to be orthonormal.
It follows from the standard theory of quadratic forms that
Lemma 3.2. A scalar product on V ( 6= ;) has an orthonormal basis.
If (ei ) is an orthonormal basis, the number of i such that g (ei ,ei ) < 0 is indepen-dent of the choice of orthonormal basis (cf Sylvester’s law of inertia on the signatureof quadratic forms).
The signature of g is typically written −+ ..+ or +..+−, ... (each minus sign cor-respond to an i such that g (ei ,ei ) < 0 etc..)
Definition 3.2. Let g be a (0,2) tensor field. g is said to be symmetric if for any vectorfields X and Y , g (X ,Y ) = g (Y , X ).
If gαβ are the components of g in some local basis of vector fields, then g issymmetric if and only if gαβ = gβα.
If g is a (0,2) symmetric tensor field, then at each p, gp is a symmetric bilinearform on Tp M . We say that g is non-degenerate if gp is non-degenerate at each p.Moreover, at each p, we can compute the signature of gp .
We can thus define a metric tensor as follows.
Definition 3.3. A metric tensor g on M is a symmetric non-degenerate (0,2) tensorfield on M of constant signature. (M , g ) is then called a semi-Riemannian manifold.If g is positive definite, then (M , g ) is called a Riemannian manifold, if sign g = −++..++, then (M , g ) is called a Lorentzian manifold.
The name pseudo-Riemannian manifold is also used instead of semi-Riemannianmanifold (not to be confused with sub-Riemannian).
Let (M , g ) be a semi-Riemannian manifold and (xα) a local coordinate system.Since g is non-degenerate, its matrix components gαβ is invertible and will be de-
noted gαβ.
Proposition 3.1. The components gαβ defines a (2,0) tensor field, called the inversemetric tensor.
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Proof. Let (xα) and (y i ) be two coordinate systems defined on a common open set.Let gαβ be the components of g in the first system, g ′
i j be the components of g in
the second. Similarly, we write gαβ and (g ′)i j for the components of the inverse.We have, by definition of the inverse,
gαβgβγ = δαγ , (g ′)i k (g ′)k j = δij ,
while by the transformation rule
(g ′)k j =∂xα
∂yk
∂xβ
∂y jgαβ.
Thus, we have
(g ′)i k ∂xα
∂yk
∂xβ
∂y jgαβ = δi
j
As in usual differential calculus, the Jacobian matrix(∂xβ
∂y j
)has inverse
(∂y j
∂xγ
). Thus,
we have
(g ′)i k ∂xα
∂yk
∂xβ
∂y jgαβ
∂y j
∂xγ= δi
j∂y j
∂xγ
i.e.
(g ′)i k ∂xα
∂ykgαγ = ∂y i
∂xγ
Multiply now by the inverse metric (gγρ) to obtain
(g ′)i k ∂xρ
∂yk= ∂y i
∂xγgγρ
Multiplying again by the inverse Jacobian matrix, we obtain
(g ′)i k = ∂y i
∂xγ∂yk
∂xρgγρ ,
i.e. gαβ transforms as a (2,0) tensor field. In view of Remark 2.2, it follows that thereexists a unique (2,0) tensor field of components given by gαβ.
If X is a vector field, then the map
Y ∈ Γ(M) → g (X ,Y )
is F (M)-linear and therefore defines a 1-form, denoted3 [X . In components, withX = Xα∂xα , we will write
[X = Xαgαβd xβ := Xβd xβ,
i.e. the components of the corresponding 1-form are denoted by the same letter butwith the indices down.
Lemma 3.3. The map [ : Γ(M) →Λ1(M) is an F (M)-linear isomorphism.
Proof. 1. Injectivity: Note that if g (V ,Y ) = g (W,Y ), for all Y ∈ Γ(M), then V =W(g is non-degenerate).
3Traditionally called the musical notation.
15
2. Surjectivity:Let θ ∈ Λ1(M) and consider (U , (xα)) a coordinate system, gαβ the compo-
nents of g and gαβ the components of the inverse matrix of gαβ. The map]θ : p → gαβ(p)θα(p)∂xβ then defines a vector field on U . Moreover, since gαβare the components of a (2,0) tensor field and θα the components of a one-form, one easily check that gαβθα transforms like a vector field. Thus, in viewof Remark 2.2, it follows that ]θ can actually be defined uniquely as a globalvector field on M .
Now we have, for all X ∈ Γ(X ), g (]θ, X ) = θ(X ) by construction (evaluate ev-erything in the (xα) coordinate system. )
From the above proof, for a one-form ν, ]ν is given by the map
]ν : θ ∈Λ(M) → g−1(θ,ν)
and by conventionνα := ]να = νβgαβ.
The raising and lowering of indices can then be generalized to arbitrary tensors inthe natural way.
Definition 3.4. Let (M , g ) be a semi-Riemannian manifold.
• Let X ∈ Tp M, then X is said to be spacelike if gp (X , X ) > 0 or X = 0, timelike ifgp (X , X ) < 0 and null if X 6= 0 and gp (X , X ) = 0.
• The set of all null vectors in Tp M is called the null cone at p ∈ M and the unionof all null and timelike vectors is the set of all causal vectors. If α is a smoothcurve, then α is said to be timelike (respectively spacelike, null, causal) if all itstangent vectors are timelike (respectively spacelike, null, causal).
Remark 3.1. One of the postulates of General Relativity is that a particle of mass mmoves along a timelike curve if m > 0 and along a null curve if m = 0. This is thegeometric analogue of the postulate that massive bodies move with a speed strictlyless than the speed of light and that massless bodies move with the speed of light.
Exercise 3.2. Extend the above definition from curves to submanifolds of M (What isa spacelike hypersurface?).
The line element
The metric g can be written locally as
g = gαβd xα⊗d xβ.
By polarization, g can of course be reconstructed from the map q(v) : v ∈ Tp M →g (v, v). The map q is called the line element of the metric g and is typically denotedd s2 with
d s2 = gαβd xαd xβ.
See for instance the explanations p56 in [O’N83].
16
3.2 Examples and basic constructions
3.2.1 Riemannian case
The simplest example of a Riemannian manifold is (Rn ,δ), where δ is the Euclideanmetric, of components δi j in the canonical coordinates of Rn .
Submanifolds of a Riemannian manifold
Lemma 3.4. Let Σ be a submanifold of a Riemannian manifold (M , g ) and j :Σ→ Mthe inclusion map. Then, the pull-back of g , j?(g ) is a Riemannian metric on Σ.
Proof. It follows from the definition that j?(g ) is a (0,2) (symmetric) tensor. More-over, given vectors X and Y in TpΣ, p ∈Σ, we have by definition
j?(g )p (X ,Y ) = gp (d jp (X ),d jp (Y )).
Since Σ is a submanifold, the differential d jp is injective at each p, thus d jp (X ) = 0if and only if X = 0. It then follows easily that j?(g )p is positive definite.
Exercise 3.3. Consider the sphereS2 = x2+y2+z2 = 1 inR3. Recall thatS2 endowedwith the induced topology is a submanifold of R3. We consider on S2 a local chart ξgiven by the coordinates (θ,φ) defined on (0,π)× (0,2π) by the inverse of the map
(θ,φ) → (cosθ, sinθcosφ, sinθ sinφ) ∈S2.
We consider on R3 spherical coordinates χ= (r,θ,φ) defined on (0,+∞)×(0,π)×(0,2π) by the inverse of the map (here by a small abuse of language, we are usingagain the label φ and θ)
(r,θ,φ) → (r cosθ,r sinθcosφ,r sinθ sinφ).
• Check that in the (r,θ,φ) coordinate system
δ= dr ⊗dr + r 2 (sin2(θ)dφ⊗dφ+dθ⊗dθ
).
and that the map χ j ξ−1 is given by
χ j ξ−1 : (θ,φ) → (1,θ,φ).
• Check that the induced metric on S2 is given by
σS2 = sin2(θ)dφ⊗dφ+dθ⊗dθ.
3.2.2 The Lorentzian case
Minkowski space
Minkowski space is the Lorentzian manifold (Rn+1,η), with η the Lorentzian metricsuch that if (xα) = (t = x0, xi ) are canonical coordinates on Rn+1, then the ∂xα areorthonormal with4 g00 =−1. Minkowski space plays the role of the Euclidean space
4Here are considering units (i.e. a specific choice of inertial frame) such that the speed of light in thevacuum is 1. In general, we have g00 =−c2 in an inertial frame.
17
in Lorentzian geometry, in the sense that it is the trivial example of a Lorentzianmanifold.
Contrary to the Riemannian case, if N is a submanifold of a Lorentzian manifold(M , g ) (such as the Minkowski space) with inclusion map j , the pull-back tensorj?(g ) does not need to define a Lorentzian metric, in fact, it does not need to definea metric at all.
Exercise 3.4. In (Rn+1,η) with coordinates (t , x, y, z), consider the hypersurfaces givenby the equations t = 0, x = 0, t 2 = x2+ y2+z2+1 with t ≥
√x2 + y2 + z2 +1, and t = x
and compute the pull back of the metric in each of these cases. In which cases do wehave a Riemannian or Lorentzian metric ?
By a small abuse of language, given P a submanifold of a semi-Riemannian man-ifold (M , g ) with inclusion map p, the pull-back tensor j?(g ) will sometimes becalled induced metric, even when it does not define a metric.
3.2.3 Products
Let M and N be semi-Riemannian manifold with metric gM and gN respectively.Consider the product manifold M ×N and let π : M ×N and σ : M ×N be the projec-tion on the first and second components. Then,
g :=π?(gM )+σ?(gN )
is a metric on g .For instance, if (N , gN ) is a Riemannian manifold, then we can construct a Lorentzian
manifold on the product Rt ×N by considering the metric
g =π?(−d t ⊗d t )+σ?(gN ).
(The index t on Rt simply expresses that t is a global coordinate on R.) Often, we’lldrop the pull back of the projections in the above formula and simply write
g =−d t ⊗d t + gN .
3.2.4 Uniformly Lorentzian metrics on Rn+1
In this section, we consider Rn+1 with canonical coordinates (x0, xi ). Latin indices1 ≤ i , j ≤ n will always refer to the spatial directions, so that for instance, if T is a ten-sor field, Ti j denotes all the components of T associated to the ∂xi , 1 ≤ i ≤ n. Greekindices will be used to denote all the components (including the ones associated to∂x0 ) of tensors.
In elliptic pde, a standard definition is that of a uniformly elliptic symbol. Thiswould correspond geometrically5 to a Riemannian metric h and a global chart6 suchthat, for some a,b > 0, we have
aXi X i ≤ hp (X , X ) ≤ bXi X i ,
for all tangent vectors X and all points p.
5This correspondance only make sense for second order elliptic pde.6One can easily remove this constraint provided we have another reference metric.
18
The aim of this section is to introduce a (naive) analogous notion for Lorentzianmanifolds. We will also provide some formulae to compute or estimate the compo-nents gαβ in terms of those of gαβ.
We start with some linear algebra.
Lemma 3.5. Let g ∈ Mn+1(R) be a symmetric matrix such that g00 < 0 and h = (gi j ),the submatrix obtained by removing the first row and the first column, is positivedefinite. Then g has sign −+ ...+ .
Remark 3.2. The point of the lemma is that we do not need to know the values of gon the whole first row and the whole first column to determine the signature of g inthe case, only the sign of g00.
Proof. Let O be an orthogonal matrix diagonalizing h and let k = M tO g MO , where
MO =(
1 00 O
).
Consider the submatrix l = (ki j ) obtained similarly by removing the first row andthe first column of k. Then, l is diagonal and
l =Ot hO = di ag (λ1, ..,λn),
with λi > 0 for all i . Furthermore k00 = g00 < 0 and the eigenvalues of g and k coin-cides. We compute the determinant of k −λI to get
p(λ) = det(k −λI ) =(
k00 −λ−k2
01
λ1 −λ− ..− k2
0n
λn −λ
)(λ1 −λ)..(λn −λ).
Let us define
f (λ) = k00 −λ−k2
01
λ1 −λ− ..− k2
0n
λn −λand its derivative
f ′(λ) =−1− h201
(λ1 −λ)2 − ..− h20n
(λn −λ)2 .
Let λm be the smallest of the λi . On (−∞,λm), f is smooth and f ′ < 0. Moreover,f (−∞) =+∞ and f (0) < 0. Thus, there is a unique negative value of λ ∈ (−∞,0), sayλ0 for which f (λ) = 0 which has to be an eigenvalue of g . Moreover, p ′(λ0) 6= 0, i.e.it is a root with multiplicity one. Since f ′ < 0 on (−∞,λm), there can be no othereigenvalue in this interval. It follows that the remaining eigenvalues must be strictlypositive.
Alternatively, from the assumptions, g restricted to R(1, ..,0) is definite negativewhile g restricted to the span of the ei = (0, ..,1, ..,0), 1 ≤ i ≤ n is definite positive. Soit follows from basic linear algebra that g has one negative eigenvalue and n positiveones.
Definition 3.5. Let g be a smooth symmetric (0,2) tensor field on Rn+1. We say thatg is uniformly Lorentzian in the coordinate system (x0, .., xn) if there exists constantsa,b > 0 such that g00 <−a, (gi j ) > b and the components of g are uniformly bounded.
19
Of course, it follows from the previous lemma that such a uniform Lorentzianmetric is in particular Lorentzian.
Given a uniformly Lorentzian metric, it will be useful (for instance, when wewant to prove estimates for solutions of wave equations) to have estimates control-ling the inverse metric as well.
Let g be a matrix as in Lemma 3.5. Let h be the submatrix of g given in com-ponents by gi j where we have removed the first row and the first column. FromLemma 3.5, we know in particular that g is invertible. Let gµν be the componentsof the inverse matrix of g and let H to denote the submatrix given in componentsby g i j obtained by removing the first row and the first column (note that in generalH 6= h−1 ).
Finally, we will denote the vector of components g0i by s(g ) (s is sometimescalled the shift vector).
Lemma 3.6. Let g be a matrix as in Lemma 3.5, i.e. g00 < 0 and h > 0, where h is the3×3 submatrix defined by hi j = gi j , 1 ≤ i , j ≤ n. Then,
•
g 00 = 1
g00 −d 2 ,
where d 2 = (h)−1(s(g ), s(g )).
• H = (g i j ) is positive definite, with
g00
g00 −d 2 h−1 ≤ H ≤ h−1.
• The new shift vector is given by
s(g−1) = 1
d 2 − g00h−1s(g ).
Proof. Let A be the square root of h−1, i.e. the positive definite symmetric matrixsuch that A2 = h−1. Then At h A = I d . Let k = M t
A g MA , where
MA =(
1 00 A
).
Then, k00 = g00, the submatrix (ki j ) obtained by erasing the first row and the firstcolumn is simply the identity matrix and s(k), the new shift vector, is given by At s(g ) =As(g ). Let B be an orthogonal matrix such that B t At s(g ) = |At s(g )|e1, where | . | de-notes the Euclidean norm and e1 = (1, ..,0) is the first vector of the canonical basis ofRn . Note that
|At s(g )|2 = [(At s(g ))t .At s(g )]1/2
= [s(g )A At s(g )]1/2 = [s(g )A2s(g )]1/2 = [h−1(s(g ), s(g ))]1/2 = d .
Consider ρ = M tB kMB , where MB is defined similarly to MA . Then, ρ00 = g00, the
submatrix (ρi j ) obtained by removing the first column and the first row is simply theidentity and s(ρ) = de1. Note that the inverse of the 2×2 submatrix with componentsρµν, 0 ≤µ,ν≤ 1 is given by
1
g00 −d 2
(1 −d−d g00
).
20
Since g−1 = MA MBρ−1M t
B M tA and the matrices MA and MB preserves the 00 com-
ponents of a matrix, we obtain the first claim.Moreover, let Hρ = (ρi j ) be the submatrix obtained by removing the first column
and the first row of ρ−1. Then,
H = AB HρB t At .
We have, for any w 6= 0, denoting ⟨ , ⟩ the usual scalar product on Rn ,
H(w, w)
h−1(w, w)= ⟨AB HρB t At w, w⟩
⟨Aw, Aw⟩
= ⟨HρB t At w,B t At w⟩⟨B t At w,B t At w⟩ ,
using that B is orthogonal and A symmetric. The second claim then follows sinceHρ is a diagonal matrix with all diagonal element 1 expect for the first which is givenby g00
g00−d 2 ≤ 1. We leave the last claim as an exercice.
3.2.5 A first look at the Schwarzshild spacetime
The Schwarzschild spacetime is one of the most important example of Lorentzianmanifold. It was discovered in 1915 by Schwarzschild as an explicit solution to theEinstein equations. It is not just one Lorentzian manifold, but a one parameter fam-ily of Lorentzian manifolds, indexed by a positive7 number m > 0 (referred to as themass). Let thus m > 0 and consider the product manifold
Mext =Rt × (2m,+∞)×S2.
The projection on the second variable will be denoted by r . r is just a smooth func-tion on our manifold taking values in (2m,+∞). On Mext , we consider the followingLorentzian metric
gext =−(1− 2m
r
)d t ⊗d t +
(1− 2m
r
)−1
dr ⊗dr + r 2σS2 ,
where σS2 denotes the pull back on Mext of the usual round metric on S2. On Mext ,we can consider local coordinates of the form (t ,r,θ,φ). Note that the above expres-sion for the metric g becomes singular as r → 2m.
We consider a change of coordinates of the form (t ,r,θ,φ) → (t + f (r ),r,θ,φ
),
where f is a function which will be determined. Our aim will be to choose f suchthat in the new coordinate system, the metric will be regular at r = 2m.
Let us thus define v(t ,r ) = t + f (r ). We have d v = d t + f ′(r )dr and thus d t =d v − f ′dr . The metric in the (v,r,θ,φ) coordinates can therefore be written as
gext = −(1− 2m
r
)(d v − f ′dr
)⊗ (d v − f ′dr
)+(1− 2m
r
)−1
dr ⊗dr + r 2σS2
= −(1− 2m
r
)d v ⊗d v +
(1− 2m
r
)f ′ (d v ⊗dr +dr ⊗d v)
−(1− 2m
r
)(f ′)2 dr ⊗dr +
(1− 2m
r
)−1
dr ⊗dr + r 2σS2 .
7Schwarzschildean spacetimes with m < 0 can also be defined, but their physical interpretation isdifferent (naked singularity vs. black hole).
21
In order to regularize the metric and cancel the terms which becomes singular atr = 2m, we see that we need
( f ′(r ))2 =(1− 2m
r
)−2
,
i.e.
f ′(r ) =±(1− 2m
r
)−1
.
This can be integrated to give us
f (r ) =±r ±2m ln(r −2M)+C , C ∈R.
Let us consider the + sign, fix C = 0 and thus define v(t ,r ) := t + r +2m ln(r −2m).The metric then takes the form
gext = −(1− 2m
r
)d v ⊗d v +d v ⊗dr +dr ⊗d v + r 2σS2 . (5)
Note that we have only done a change of coordinates. In particular, for r > 2m, ourmetric is still Lorentzian, since this is independant of our choice of coordinates, butin fact the above expression defines a Lorentzian metric for any r > 0, since for any
α ∈R, the matrix
(α 11 0
)has signature −+ .
We consider thus another manifold given by M = Rv × (0,+∞)×S2. We can en-dow this manifold with the Lorentzian metric given by the expression found in (5),which we denote by gM . The map
ψ : Mext → M
(t ,r,ω) → (v(t ,r ),r,ω)
then provides a embedding of Mext into M which preserves the metric ψ∗(gM ) =gMext . Note that ψ(Mext ) =Rv × (2m,+∞)×S2 is strictly smaller than M .
Remark 3.3. M is still not the largest regular Lorentzian manifold than "contains"Mext . First, we could have defined an extension of Mext by taking the minus signinstead of the plus sign. The two differents extensions can in fact be glued together,corresponding to an exterior region, a black hole region (the one corresponding to theinterior of M \ψ(Mext ) and a white hole region ( a similar region obtained by takingthe minus sign above). This manifold can in fact still be extended !!
3.3 The Levi-Civita connection
Given a semi-Riemannian manifold, we are looking for an operation replacing theusual notion of derivative for tensorial objects. We start by axiomatizing the keyproperties that we would like.
Definition 3.6. A connection D on a smooth manifold M is map of the form
D : Γ(M)×Γ(M) → Γ(M)
(V ,W ) → DV W
such that DV W is F (M)-linear in V , R-linear in W and the Leibniz rule is verified
DV ( f W ) =V ( f ).W + f DV W,
for all f ∈F (M).
22
DV W is called the covariant derivative of W with respect to V (for the connectionD).
Let D be a connection on M . Let p ∈ M and consider v ∈ Tp M . Let V be anyvector field such that Vp = v . Then, for any W ∈ Γ(M), (DV W )(p) is independent ofthe choice of V (it depends only on v). Thus, for each p, we can define map (stilldenoted D)
D : Tp M ×Γ(M) → Tp M
(v,W ) → Dv W.
Given a semi-Riemannian manifold, there can be plenty of connections, but there isa unique one that satisfies the two extra conditions written below8
Theorem 3.1. Let (M , g ) be a semi-Riemannian manifold. Then, there exists a uniqueconnection D such that
1. D is torsion free: [V ,W ] = DV W −DW V ,
2. X (g (V ,W )) = g (DX V ,W )+ g (V ,DX W ).
D is called the Levi-Civita connection of (M , g ) and is characterized by the Koszulformula
2g (DV W, X ) =V g (W, X )+W g (X ,V )−X g (V ,W )−g (V , [W, X ])+g (W, [X ,V ])+g (X , [V ,W ]).
Remark 3.4. The first condition can be interpreted as a twist or torsion free condi-tion. Without it, geodesics (which we are going to define soon) would not completelydetermined the connection. The second condition means essentially that Dg = 0, i.e.the metric is constant with respect to this notion of derivative.
Proof. Let D be a connection satisfying the above two conditions. Inserting the con-ditions on the RHS of the Koszul formula, it follows that it must hold. Then the non-degeneracy of the metric implies that DV W is uniquely determined. Reciproquely,let F (V ,W, X ) be given by the RHS of the Koszul formula. Let V ,W ∈ Γ(M) and checkthat X → F (V ,W, X ) is F (M)-linear. Thus, the formula defines a one-form and wedenote it image by ] as 2DV W (i.e. 2g (DV W, X ) = F (V ,W, X ) ). Then the Koszulformula holds and from it we can check that all the conditions hold. For instance,
g (X , [V ,W ]) = 2g (DV W, X )−V g (W, X )−W g (X ,V )+X g (V ,W )+g (V , [W, X ])+g (W, [V , X ])
and similarly (exchanging V and W )
g (X , [V ,W ]) =−2g (DW V , X )+V g (W, X )+W g (X ,V )−X g (V ,W )−g (V , [W, X ])−g (W, [V , X ])
Adding the equations give the torsion free property.
8To motivate, naively and quickly, the definition below, one can imagine a submanifold P ofRn with itsinduced metric. Consider a point p ∈ P and an orthnormal frame in Tp P at that point. Consider movingthat frame along a curve on P passing through p such that the moving frame stays parallel to the originalone. Then, we would like that our notion of derivative to be such that the derivative of the moving framein the direction of the tangent vector of the curve be zero. From this constraint, one can reconstruct thedefinition given below.
23
Let us also check that with this definition, D is indeed F (M)-linear in the firstvariable. Let f ∈F (M). Then,
2g (D f V W, X ) = f V g (W, X )+W g (X , f V )−X g ( f V ,W )− g ( f V , [W, X ])+ g (W, [X , f V ])+ g (X , [ f V ,W ])
= f V g (W, X )+W ( f g (X ,V ))−X ( f g (V ,W ))− f g (V , [W, X ])
+ g(W, X ( f )V + f [X ,V ]
)+ g(X , f [V ,W ]−W ( f )V
)= f
(V g (W, X )+W g (X ,V )−X g (V ,W )− g (V , [W, X ])+ g (W, [X ,V ])+ g (X , [V ,W ])
)= f 2g (DV W, X ).
Exercise 3.5. Check the other conditions and finish the proof.
From the Koszul formula, it follows that, for any p, DV W (p) only depends on thevalue of W in a neighborhood of p (since it only depends on W (p) and its derivativeat p. Thus, given any open set U ⊂ M , we can consider D defined on Γ(U )×Γ(U ).
Definition 3.7. Let (xα) be a coordinate system. The Christoffel symbols are the realvalued functions Γα
βγsuch that
D∂xβ∂xγ = Γαβγ∂xα .
Lemma 3.7. We have Γαβγ
= 12 gαρ
(∂xβgγρ +∂xγgβρ −∂xρ gβγ
).
In particular, Γαβγ
= Γαγβ
.
Proof. We have,g (D∂
xβ∂xγ ,∂xρ ) = Γαβγgρα.
On the other hand, using the Koszul formula, all the commutators vanish and
2g (D∂xβ∂xγ ,∂xρ ) = ∂xβgγρ +∂xγgβρ −∂xρ gβγ.
Conclude by inverting gρα in the first equation.
Exercise 3.6. 1. Let (S2,σS2 ) be the usual 2-sphere endowed with the round met-ric. Compute its Christoffel symbols in the (θ,φ) coordinate system.
2. Let (R3+1,η) be the Minkowski space and consider spherical coordinates so thatthe line element takes the form
d s2 =−d t 2 +dr 2 + r 2dσS2 .
Compute its Christoffel symbols.
Of course, for the Euclidean or the Minkowski space in Cartesian coordinates9,the covariant derivatives in the direction of ∂xα just reduces to partial derivativesand the Christoffel symbols vanish.
We now seek to extend our connection as a derivation for arbitrary tensors. Firsta definition.
9We call Cartesian coordinates, global coordinates onRn orRn+1 such that the metric components aregiven by the matrices di ag (1,1..,1) in the Euclidean case and di ag (−1,1, ..1) in the Minkowskian case.
24
Definition 3.8. A tensor derivation D is a set of maps D(= Drs ) : T r
s (M) → T rs (M),
where T rs (M) is the set of tensor fields of type (r, s), such that,
1. for a tensor field A⊗B, D(A⊗B) = D A⊗B + A⊗DB,
2. D(C A) =C (D A) for any contraction C .
We have easily.
Proposition 3.2. Given a vector field V and an R-linear map δ : Γ(M) → Γ(M) suchthat
δ( f X ) =V ( f )X + f δ(X ),
there exists a unique tensor derivation D on M such that D00 =V and D1
0 = δ.
Thus, the Levi-Civita connection extends naturally to arbitrary tensor fields.
Proof. D00 (derivation of functions) and D1
0 derivation of vector fields are given. As-sume that such a general D exists. Let θ be a one-form and X a vector field. Then,θ⊗X is a (1,1) tensor. From the contraction rule of a derivation, we must have
D(C (θ⊗X )) =C (D(θ⊗X )).
Using the rule for the derivation of a tensor product,
D(θ⊗X ) = Dθ⊗X +θ⊗D X .
Using the definition of the contraction, this gives
D(θ(X )) = Dθ(X )+θ(D X )
and thusDθ(X ) =V (θ(X ))−θ(δX ),
which determines uniquely Dθ. Moreover, one easily check that the above expres-sion for Dθ is F (M)-linear and thus defines a one-form, since for any function f
V (θ( f X ))−θ(δ( f X )) = V(
f θ(X ))−θ (
V ( f )X + f δ(X ))
= V ( f )θ(X )+ f V (θ(X ))−V ( f )θ(X )− f θ(δ(X ))
= f (V (θ(X ))−θ (δ(X ))) .
For an arbitrary tensor A of type (r, s), we have similarly,
D A(θ1, ..,θr , X1, .., Xs ) = V (A(θ1, ..,θr , X1, .., Xs ))−r∑
i=1A(θ1, ..,Dθi , ..,θr , X1, .., Xs )
−s∑
j=1A(θ1, ..,θr , X1, ...,δX j , ..., Xs ).
Thus D is uniquely determined. One then needs to check that the above formulaindeed defines a tensor derivation. The fact that D distributes on tensor productsis easy from the above formula. For the contraction property, let us prove that D1
1commutes with C . We have trivially that D commutes with C for tensor productθ⊗X . The result then follows since any (1,1) tensor can be written as a sum of suchterms.
25
Exercise 3.7. 1. Recall that given two vector fields X ,Y ∈ Γ(M), [X ,Y ] is a vectorfield. Show that it defines a tensor derivation called the Lie derivative (notationLX Y = [X ,Y ]).
2. Let (xα) be a coordinate system. Compute L∂αgβγ := (L∂αg
)βγ.
3. Deduce that L can never coincide with the tensor derivation associated withthe Levi-Civita connection of a metric. (Of course, this also follows from thefact that DV W is F (M)-linear in V ).
4. Let X and Y be two vector fields and p ∈ M. Let φt denotes a local flow of Xnear p ∈ M. Prove that
limt→0
1
t
(dφ−t (Yφt (p))−Yp
)t=0 = [X ,Y ]p . (6)
Definition 3.9. Let T be any tensor field of type (r, s). Then we define DT , the covari-ant differential of T , as the (r, s +1) tensor field
Notations:Let (xα) be a coordinate system and T a tensor field of type (r, s). Then the covariantderivative of T in the direction of α is often denoted Dα instead of D∂αx
.Moreover, the component of the covariant derivative DαT are sometimes written
asT i1,..,ir
j1,.., js ;α.
The ";" should be compared with the usual "," used for instance often in PDEs nota-tions to denote partial differentiations.
Exercise 3.8. Prove that Dg = 0.
3.4 Isometries
Definition 3.10. Let (M , g ) and (N ,h) be semi-Riemannian manifolds. An isometryfrom M to N is a diffeomorphism φ : M → N such that
φ?(h) = g .
This means thathφ(p)
(dφp (v),dφp (w)
)= gp (v, w),
for all v, w ∈ Tp M , p ∈ M .Recall the definition of the flow φ of vector field X ∈ Γ(M), cf Definition 2.15.
If the flow is complete, then for every s ∈ R, φs is a diffeomorphism from M to M .Similarly, if the flow is non-complete, its flow will define diffeomorphisms of theform, φs : U →φs (U ).
A natural question is: which vector field X have a flow generating isometries ?
Proposition 3.3. Let X ∈ Γ(M) and A ∈ T 0s (M) (a tensor field of type (0, s)). Consider
an open set D ⊂ M on which the flow of X , ψs , is well defined for s sufficiently small.Then, in D,
LX A = limt→0
1
t
(ψ?t (A)− A
),
where LX A denotes the Lie derivative of A with respect to the vector field X .
26
Proof. We do the proof in the case of a (0,2) tensor field A and we will use formula(6) from above.
Since LX is a tensor derivation, for vector fields V and W ,
+ Aψt (p)(Vψt (p),Wψt (p))− Ap (Vp ,Wp ) = I + I I .
Let α be the integral curve of X starting at p, i.e α(t ) =ψt (p), α(0) =ψ0(p) = p.Then, by definition,
limt→0
1
tI I = d
d t(Aα(Vα,Wα))|t=0 = Xp (A(V ,W )).
For I , we write first
I = A(dψt (Vp )−Vψt (p),dψt (Wp )
)+ A(Vψt (p),dψt (Wp )−Wψt (p)
).
Since ψt ψ−t = I d ,
dψt (Vp )−Vψt (p) = dψt (Vp −dψ−t Vψt (p)),
and similarly for the second term. The result then follows from (6) and the fact thatdψt → I d as t → 0.
Definition 3.11. A Killing vector field on a semi-Riemannian manifold is a vectorfield X for which the Lie derivative of the metric tensor vanishes LX g = 0.
Proposition 3.4. Let X be a vector field and denote its local flow by ψt . X is Killing ifand only if all its (local10) flows are isometries.
Proof. Let p ∈ M and U a neighborhood of p on which ψt is well defined for all tsufficiently small. Then, if all ψt are isometries ψ?t (g ) = g and thus, by the aboveLX (g ) = 0. Reciprocally, let LX g = 0 and let ψt be a local flow of X . By standardproperties of ψt , we have ψsψt =ψs+t for s, t small enough. The previous proposi-tion then implies that, for v, w tangent vectors at some point in the domain of theflow and s sufficiently small
limt→0
1
t
(g (dψs+t (v),dψs+t (v)− g (dψs (v),dψs (w)
)) = 0
Thus, the real-valued functions s → g (dψs (v),dψs (w)) has zero derivative and isthus constant, i.e.
g (dψs (v),dψs (w)) = g (v, w),
which implies that ψs is an isometry.
10A Killing vector field on a geodesically complete semi-Riemannian manifold is in fact necessarilycomplete, cf [O’N83], page 254.
27
Definition 3.12. For any vector field X , define its deformation tensor as
πX (V ,W ) = 1
2
(g (DV X ,W )+ g (V ,DW X )
).
From the definition, πX is indeed F (M)-linear in each of its two arguments andthus a (0,2) tensor field. In coordinates
Xπαβ = 1
2
(DαXβ+DβXα
):= D(αXβ),
where for any tensor Aαβ, A(αβ) denotes its symmetrized version
A(αβ) := 1
2
(Aαβ+ Aβα
).
Lemma 3.8.2Xπ=LX g .
Exercise 3.9. 1. Prove Lemma 3.8.
2. Consider Minkowski space with Cartesian coordinates (xα) = (t , xi ). Show thatthe translations ∂xα , the rotations xi∂x j − x j∂xi , and the hyperbolic rotationst∂xi +xi∂t are all Killing fields.
3. Prove that if φ : M → N is an isometry, then dφ(DX Y ) = Ddφ(X )(dφY ), for allX ,Y ∈ Γ(M). (Recall first what dφ(X ) is well defined for X a smooth vector fieldand φ a diffeomorphism ).
3.5 The induced connection on a curve, parallel translations andgeodesics
Definition 3.13. Let N and M be smooth manifolds and φ : N → M a smooth map. Avector field on φ is a map Z : N → T M such that π Z = φ, where π is the canonicalprojection π : T M → M.
In other words, for any q ∈ N , Z (q) = (φ(q), v), where v ∈ Tφ(q).In terms of vector bundles, Z is thus a section of the pullback bundle φ?(T M).Consider now a curve α : I → M . A vector field on α is according to the above
definition a map that assigns to each s ∈ I a tangent vector at α(s). Let Γ(α) denotethe set of all vector fields along α.
A particular example is given by the tangent vector α′ to α itself. Another one isobtained from any vector field V ∈ Γ(M), as Vα : t →Vα(t ).
Proposition 3.5. Let α : I → M be a curve. Then, there is a unique map
d
d s: Γ(α) → Γ(α)
Z → Z ′ = d Z
d s
called the induced covariant derivative such that
28
1. dd s is R-linear.
2. Leibniz rule: (hZ )′ = dhd s Z +hZ ′, for any h ∈F (I ),
3. (Vα)′(t ) = Dα′(t )(V ), ∀V ∈ Γ(M).
Moreover, we then have, for any vector fields Z1 and Z2 on α,
d
d tg (Z1, Z2) = g (Z ′
1, Z2)+ g (Z1, Z ′2).
Proof. 1. Uniqueness: This follows from the first three properties only. Let usfirst assume that α lies in a a single coordinate patch (U , (xi )).
For any Z ∈ Γ(α), we can write
Z (t ) = Z i (t )∂xi ,
where Z i (t ) = Z (t )(xi ).
By the first two properties,
d Z
d s= (Z i )′∂xi +Z i [
(∂xi )|α]′ .
Note that by the usual bump function argument, it follows that if V ∈ Γ(U ),then the third property still holds, i.e.
(Vα)′(t ) = Dα′(t )(V ), ∀V ∈ Γ(U ).
(Indeed, given t0 ∈ I , consider a bump function f around α(t0), then f V canbe extended to a global vector field on M , use then property 3, expands usingproperty 2 and evaluate everything at t0.)
Thus by the third property,[(∂xi )|α
]′ (t ) = Dα′(t )∂xi .
In particular, the connection completely determined d Zd s .
In the general case (i.e. α not restricted to a single coordinate patch), let J bean non-empty open subinterval on which α(J ) lies in a a single coordinatepatch (U , xi ). Let I d
d s the operation associated to the original α and J dd s the
one associated to α|J . From the above, J dd s is already uniquely determined.
Let us show that given any Z ∈ Γ(α),[I d
d sZ
]|J= J d
d sZ|J .
Let t0 ∈ J , f be a bump function around α(t0) with support in U and h be abump function (defined on R ) around t0 with support in J . As before we canwrite
Z (t ) = Z i (t )∂xi
for t ∈ J . We consider hZ i and f ∂xi and extends them smoothly by 0 outsideof respectively J and U .
29
Then, on one hand,
(Z i h f ∂xi )′ = (h · f α ·Z )′ = (h f α)′Z +h f αZ ′,
and thus (Z i h f ∂xi )′(t0) = Z ′(t0). On the other hand,
(Z i h f ∂xi )′ = (Z i h)′ f ∂xi +hZ i ( f ∂xi )|α)′
= (Z i h)′ f ∂xi +hZ i Dα′ ( f ∂xi ).
We then evaluate this at t0 and recognise on the right-hand side the expression
previously found for[
J dd s Z|J
].
2. Existence: on any subinterval J of I such that α(J ) lies in a coordinate neigh-borhood, define Z ′ be the above formula. One can then check that all fourproperties hold (with I replaced by J ). Let J1 and J2 be two such subintervalsof I , such that J1 ∩ J2 6= 0. The above formula gives two vector fields on J1 ∩ J2,called them Z ′
1 and Z ′2. Since the first three properties hold, we can apply our
uniqueness statement and thus we must have Z ′1 = Z ′
2 on J1 ∩ J2 6= 0. In otherwords, the above formula defines a single vector field on α.
Exercise 3.10. Check indeed that all four properties hold as claimed in the proof. (Forthe last property, check that it first hold if α lies in a single coordinate neighborhoodfor Z1 and Z2 given by [∂xi ]α, [∂x j ]α.)
From the above proof, we obtain the coordinate expression
Z ′ = d Zβ
d s
(∂xβ
)|α+ZβDα′ (∂β)
= d Zβ
d s
(∂xβ
)|α+Zβ(xρ(α))′Γγ
βρα (∂xγ )|α ,
where Zβ := Z (xβ) : I →R are the components of Z along α.Recall that a special case of a vector field along α is given by the tangent vector
field to α, α′. Its derivative along α, α′′ is sometimes called the acceleration of thecurve α. (Note that the first ′ does not depends on the geometry, but the seconddoes.)
A vector field Z on α such that Z ′ = 0 is said to be parallel. The above computa-tion leads to
Proposition 3.6. Let α : I → M be a curve, a ∈ I and z ∈ Tα(a)M. Then, there exists aunique parallel vector field Z on α such that Z (a) = z.
Proof. Let J be a non-empty open subinterval of I containing a and such that α(J )is contained on a unique coordinate chart. From the above, we must have
0 = d Zβ
d s+ (xρ(α))′Γγ
βρZβ.
This is a system of linear odes which therefore has a unique solution defined on thewhole interval J for given initial data.
Now, if J1 is another interval such that α(J1) is contained in a unique coordinatechart and J1 ∩ J 6= 0, let then t1 ∈ J1 ∩ J 6= 0 and z(t1) = Z (t1), where Z (t1) has been
30
obtained uniquely by solving the equation on J . By the same argument, Z|J1 is thengiven uniquely by solving the equations in coordinates with the data given by z(t1).Since I can written as a union of such subintervals, we are done. (More precisely,given any t1 ∈ I , α([a, t1]) can be covered by a finite number of coordinate neighbor-hoods. Using the uniqueness statement and the finite of neighborhoods, Z can bedefined on on the whole of [a, t1] uniquely. Since this holds for any t1, the statementof the proposition holds.)
Definition 3.14. With the above notation, given a,b ∈ I the map sending z ∈ Tα(a)Mto Z (b) ∈ Tα(b)M such that Z ′ = 0 and Z (a) = z is called a parallel translation.
Exercise 3.11. Prove that parallel translation is a linear isometry from (Tα(a)M , gα(a))to Tα(b)M , gα(b)).
We can now define the notion of geodesics.
Definition 3.15. A geodesic is a curve α such that α′′ = 0.
Lemma 3.9. Letα(s) be given in local coordinates as (xα(s)). Then the geodesic equa-tions read
(xρ)′′+Γρβγ
(xβ)′(xγ)′ = 0.
The geodesic equation is therefore a system of second order odes. Since it is non-linear, (due to the product of x ′ in the second term), it does not need to be definedglobally, but given initial data it always has a unique maximal solution.
More precisely, we have by standard application of the Cauchy-Lipschitz theo-rem,
Lemma 3.10. Let v ∈ Tp M. Then there exists a unique maximal solution γ : I → Mto the geodesic equation such that γ′(0) = v where is I is an open interval and 0 ∈ I .(Note that since v ∈ Tp M, γ(0) = p. )
Such a maximal geodesic (what does maximality refer to here ? ) is called a max-imal geodesic or inextendible geodesic.
Definition 3.16. We say that (M , g ) is geodesically complete if every inextendiblegeodesic is defined on the whole of R (i.e the geodesic equation in (M , g ) always admitglobal solutions).
Lemma 3.11. Let a 6= 0, τ : s → as + b be a straight line and J = τ−1(I ). Then, ifγ : I → M is a geodesic, so is γτ : J → M .
Exercise 3.12. what about other changes of parametrization?
We will note (γv , Iv ) the geodesic associated to the vector v as above.
Note that any constant curve is a geodesic.
Exercise 3.13. • Show that the (non-constant) geodesics of the Euclidean or Minkowskispace are the straight lines.
31
• A curve α is called a pregeodesic if it has a reparametrization as a geodesic. Letnow α be a regular curve (α′ 6= 0) such that α′′ and α′ are collinear, i.e α′′(s) =f (s)α′(s) for some f .
1. Show that β=αh for β a geodesic if and only if h′′+ (f h
)h′2 = 0. Note
that this is a non-linear ode and that it may not have global solutions onthe interval of definition of α.
2. Assume now that g (α′,α′) 6= 0 and consider a constant speed reparametriza-tion, i.e. such that g (β′,β′) = c for some c ∈ R. Prove that any such con-stant speed reparametrization of α is a geodesic.
3. Prove that either g (α′,α′) = 0 identically on α, or that it never vanishes.
4. Deduce from the above that if g (α′,α′)(s0) 6= 0 for some s0, then α is apregeodesic.
5. The above proof does not work is g (α′,α′) = 0 identically on α but theresult still holds. To prove it, assume that β = α h is a geodesic with ha diffeomorphism such that h′ > 0 and write α = βk, with k = h−1 andderive the equation (logk ′)′ = f . This can then be solved. )
• Consider R3 with its usual coordinates (x, y, z).
1. Prove that there exists a unique connection D verifying
D∂x (∂y ) = ∂z , D∂x (∂y ) = ∂z (7)
D∂x (∂z ) = −∂y , D∂z (∂x ) = ∂y (8)
D∂y (∂z ) = ∂x , D∂z (∂y ) =−∂x (9)
2. One of the conditions in the definition of the Levi-Civita connection is thetorsion free connexion [V ,W ] = DV W −DW v. Prove that (7)-(9) indeeddefines a unique connection but that it is not torsion free.
3. Prove that its geodesics are straight lines. This proves in particular thatgeodesics do not define uniquely a connection (They actually define uniquelya torsion free connection).
4. What is the parallel transports of (1,0,0) and (0,1,0) along the z axis ?
3.6 Physical interpretation of causal geodesics
Recall Newtown’s equation from classical mechanics. In any Galilean frame, for aparticule of mass m,
mx =∑F.
When the forces are only gravitational, then∑
F = −m∇φ, where φ is the gravita-tional potential. The equation then reduces to
x =−∇φ(x). (10)
This is a second order ode, like the geodesic equations, whose solution depends onlyon the initial position and initial velocity of the particule11
11The fact that the initial mass m appearing on the left-hand side of (??) equal the gravitational mass ap-pearing on the right-hand side is the "weak equivalence principle" of Einstein: the acceleration impartedby a body by a gravitational field is independent of the nature of the body.
32
In General Relativity, the replacement of Newton’s equation (10) is the geodesicequation. More precisely, a particular of mass m, which is free falling (no externalforces apart the effect of gravity ), moves along a geodesic whose tangent vector istimelike if m > 0, null if m = 0. These last requirements are the general relativisticrequirement that no particules can move faster than the speed of light.
Thus in General Relativity, there are no gravitational force, there is only geom-etry. To obtain a closed physical theory, it remains to have a prescription for thisgeometry. More specifically, in Newtonian mechanic, the gravitational potentialφ isobtained through Poisson equation
∆φ= 4πGρ, (11)
where ρ is the local mass density, G a constant (the gravitational constant) and ∆=∑i
∂2
∂2xi is the usual Laplacian.Thus, although we have already a replacement for Newton’s equation, we still
need to find a replacement for Poisson equation. Since g appears in the geodesicequation, we would like this replacement to be an equation on g to obtain a closedsystem.
3.7 The exponential map
Definition 3.17. Let p ∈ M. Let Up be the set of vectors in Tp M such that for v ∈ Tp M,γv is defined on [0,1]. The exponential map is then
expp : Up → M
v → γv (1).
Remark 3.5. By smooth dependance with respect to the initial data (cf Cauchy-Lipshitz),the exponential map is a smooth function on Up .
Remark 3.6. With the above notation, for any t ∈R, the map s → γv (t s) is a geodesic.Moreover, it has initial velocity t v and position p. Thus, γv (t s) = γt v (s) by unique-ness. In particular, we have
expp (t v) = γt v (1) = γv (t ).
The curve t → γv (t ) will be called a radial geodesic (emanating from p).
Exercise 3.14. Check that the Cauchy-Lipschitz theorem from standard ode theoryguarantees that Up contains a neighborhood of 0 ∈ Tp M and in fact that Up is open.
Proposition 3.7. For each p ∈ M, there exists an open neighborhood V of 0 in Tp Msuch that expp is a diffeomorphism from V to some open neighborhood U of p ∈ M.
Proof. Let (xα) be coordinates on M . Recall that the (xα) induces a global coordinatesystem12 (Xα) on Tp M given by
X = (Xα∂xα
)|p ,
for any X ∈ Tp M . Consider the differential of the exponential map at 0 ∈ Tp M . Wehave
12These coordinates are sometimes called conjugate to the (xα).
33
d expp (0) : T0Tp M → Texpp (0)M = Tp M .
Let V ∈ T0Tp M . Since Tp M is a vector space, we can identify T0Tp M with Tp M .More precisely. We can write V = V α [∂Xα ]|0∈Tp M . Then, α : s → sV α[∂xα ]|p∈M is a
curve in Tp M whose tangent vector is V . Denote by V the vector V α[∂xα ]|p∈M (the ˜map is sometimes called the canonical isomorphism in this context).
Then by definition and the above remark,
d expp (0)(V ) = d
d s
(expp (α(s))
)|s=0
= d
d s
(expp (sV
)|s=0
=[
d
d sγV (s)
]|s=0
= V .
Thus, d expp (0) coincides with the ˜ map. The result then follows by the inversefunction theorem.
Exercise 3.15. Show that the ˜ map is independent of the choice of coordinates.
Definition 3.18. If U and V are as in the preceeding proposition and if U is star-shaped around 0, then V is called a normal neighborhood of p.
Recall that U starshaped around 0 means that for any x ∈ U , sx ∈ U for all 0 ≤s ≤ 1. (Note that U ∈ Tp M is a subset of a vector space, so that sx makes sense.)
Note that if U is a neighborhood of 0, then it contains a starsphaped neighbord-hood of 0.
Proposition 3.8. If V = expp (U ) is a normal neighborhood of p ∈ M, then for eachpoint q ∈ V , there exists a unique geodesic σ : [0,1] → V from p to q. Furthermoreσ′(0) = exp−1
p (q) ∈U .
Proof. Let q ∈ V and v = exp−1p (q) ∈ U . Since U is starshaped, the ray ρ : t → t v ,
0 ≤ t ≤ 1 is contained in U . Let σ= expp ρ. Then σ is a geodesic from p to q .For the uniqueness, consider τ : [0,1] → V be any geodesic joining p to q with
values in V . Let w = τ′(0), so that τ = γw . The difficulty is that we do not know apriori that w ∈U .
Recall that by definition t → expp (t w) is defined for t such that γt w is definedon [0,1]. Since γt w (1) = γw (t ) for all t sufficiently small, it follows by uniquenessthat t → γt w (1) = γw . In particular, they have the same interval of definition, so thatγt w (1) is defined for t ∈ [0,1]. Thus, t → expp (t w) is defined for t ∈ [0,1]. Moreover,we have τ(t ) = expp (t w) for all t ∈ [0,1]. However, we still do not know that w ∈U .
Consider the set of all t ∈ [0,1], such that t w ∈ U . It is a non-empty subset of[0,1]. Since U is star-shaped around 0, E is an intervall. Since U is open, E is openin [0,1]. Let t0 be the supremum of E .
Recall that the map τ takes value in V by assumption, so that in particular, τ(t0) ∈V and thus exp−1
p τ(t0) ∈U . Now by continuity,
exp−1p τ(t0) = lim
t→t0,t<t0exp−1
p τ(t ) = limt→t0,t<t0
t w = t0w.
34
Thus, t0w ∈ U , t0 = 1, w ∈ U and q = expp (w). Since expp is a diffeomorphism onU , it follows that w = v and hence that τ=σ.
Remark 3.7. Note that if we do not specify the domain of definition of σ, then theuniqueness does not hold anymore since we can make a change of parametrization.
Exercise 3.16. Let S2 be the unit sphere in R3 endowed with its round metric.
1. Show that the rotations of S2 are isometries.
2. Let γ be a geodesic and φ : (M , gM ) → (N , gN ) an isometry. Show that φ(γ) isalso a geodesic.
3. Show that on S2, the great circles are geodesics.
4. Show that given any two distinct points onS2, they can be joined by two distinctgeodesics (where we consider two geodesics to be equal if they have same imagein S2 to account for the parametrization freedom).
Definition 3.19. A broken geodesic is a (continuous) piecewise smooth curve witheach piece being a smooth geodesic.
Recall that a piecewise smooth curve is a continuous map γ : I → M , where Iis an non-empty interval such that if I = [a,b], then there exists a finite partitiona = t0 < t1.. < tk = b such that each curve segment γ|[ti ,ti+1] is a smooth curve, whilefor an interval with open endpoints, say I = [a,b), we require that for each non-empty closed interval J ⊂ I , the restriction γ|J is piecewise smooth.
A classical result in differential geometry asserts that a manifold is connectedif and only if it is path connected. Here is a slight variation in the case of semi-Riemannian manifolds.
Lemma 3.12. A semi-Riemannian manifold M is connected if and only if two pointsof M can be joined by a broken geodesic.
Proof. Let M be connected and p ∈ M . Let C be the set of points that can be con-nected to p by broken geodesics.C is open: indeed if q ∈ C , let V be a normal neighborhood around q . Then fromthe above lemma, V ⊂C .M \C is open: Let q ∈ M \C and again V be a normal neighborhood around q . ThenV ⊂ M \C . For, if q ′ ∈ V ∩C , then from the above, there exists a geodesic from q to q ′and from p to q ′ and thus, a broken geodesic from q to p, which is a contradiction.Since C 6= ;, it follows from the connectedness of C that C = M . The converse istrivial.
Let V be a normal neighborhood of p ∈ M . Let (eα) be an orthonormal basis forTp M . For any q ∈ V , let (xα(q)) be determined by
exp−1p (q) = xα(q)eα.
In other words, given q ∈ V , we first compute vq = exp−1p (q). Then the components
of vq in the orthonormal basis eα provides the xα. The xα are smooth functionssince they can be written as a composition of smooth functions and it is also imme-diate that they form a local coordinate system near p.
35
Lemma 3.13. The map q → xα(q) determines a local coordinate system on V callednormal coordinates.
If ωα is the dual basis to the eα, then xβ expp =ωβ on U . Moreover,
Proposition 3.9. Let (xα) be normal coordinates. Then, gαβ(p) = g (p)(eα,eβ) andΓαβγ
(p) = 0.
Remark 3.8. Note that gαβ(p) = g (p)(eα,eβ) means that the matrix gαβ(p) coin-cides with the Minkowski (up to an reordering the eα) or the identity matrix in theLorentzian or Riemannian case. In physics, these coordinates are sometimes referredto as "local inertial frames" essentially because the laws of physics written in thesecoordinates will look similar (up to a first order term) to those written in Minkowskispace or the Euclidean space, at least on a sufficiently small neighborhood of p. Inother words, if your laboratory is sufficiently small13 and if you make measurementsvery quickly (we need a small region of spacetime, not just of space), then the measure-ments should converge to those obtained in say Minkowksi space, i.e. we can neglectall gravitational effects (of course, if your lab becomes too small, then quantum effectsmight appear, but this is another story).
Proof. Let v = vαeα ∈ Tp M , where eα are orthonormal as above and vα the compo-nents of v in the (eα) basis. Since expp (t v) = γv (t ) (for t small enough),
xα(γv (t )) = xα(expp (t v)) = t vα.
Hence, v = γ′v (0) = vα[∂xα ]|p , i.e. the components of v in the([∂xα ]p
)basis are also
given by vα. It follows that eα = [∂xα ]p and hence that gαβ(p) = g (p)(eα,eβ).
Moreover, from the above expression, we have d 2
d t 2
(xα(γv (t ))
) = 0, and since γv
is geodesic, this implies that
Γαβγ(γv (t ))d
d t
(xβ(γv (t ))
) d
d t
(xγ(γv (t ))
)= 0.
Evaluated at t = 0, we get that
Γαβγ(p)vβvγ = 0,
for all v , which implies, by polarization, that Γαβγ
(p) = 0.
Remark 3.9. With the above notations, the Jacobian matrix of the differential of theexponential map at p at any point v ∈U with respect to the coordinates associated tothe eα on Tp M and the normal coordinates on expp (U ) is just the identity matrix.
Exercise 3.17. Prove that if (xα) is a normal coordinate system and X = Xα∂xα withXα constants, then the integral curve of X through p is a geodesic.
13Smallness should always be measured with respect to something. Here, we want the Γ to stay smallon some coordinate patch and since at p we have arranged that they vanish, we need the size of the labto be small compared to the derivatives of the Γ. These derivatives are not very geometric, so if we wantsomething more physical (i.e. less dependent on our choices of gauge), we need a tensorial quantity atthe level of the first derivative of the Γ. This is the curvature, soon to be defined in these lectures.
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4 Curvature
4.1 The Riemann curvature tensor and its symmetries
For a surface in R3, a well known notion of curvature is that of the Gaussian cur-vature. This is an intrinsic invariant of the surface, in the sense that Gauss provedthat the Gaussian curvature is independent of the fact that the surface happens tobe embedded in R3. The generalization of the the Gaussian curvature to arbitrary(semi-)Riemannian manifold, led Riemann to Riemannian geometry.
Lemma 4.1. Let (M , g ) be a semi-Riemannian manifold and D its Levi-Civita con-nection. The function
R : Γ(M)3 → Γ(M)
(X ,Y , Z ) → RX Y Z := DX DY Z −DY DX Z −D[X ,Y ]Z
defines a (1,3) tensor field on M called the Riemann curvature tensor of (M , g ).
Remark 4.1. Eventhough R only takes 3 vector fields as arguments, it is a (1,3) tensorfield in the sense that given any one-form ω, the map
(X ,Y , Z ,ω) →ω(RX Y Z )
is a F (M) multi-linear map from Γ(M)3 ×Λ1(M) to R.
Proof. For R to define a tensor field, one needs R to be F (M) linear in each of itsarguments. Let f ∈M , then
RX , f Y Z = DX D f Y Z −D f Y DX Y −D[X , f Y ]Z
= DX(
f DY Z)− f DY DX −DX ( f )Y + f [X ,Y ]Z
= X ( f )DY Z + f DX DY Z − f DY DX −X ( f )DY Z − f D[X ,Y ]Z
= f RX ,Y Z .
The other cases are similar.
Remark 4.2. The sign of R is conventional, i.e. some authors, for instance O’Neil[O’N83]) define R as above but with an extra minus sign everywhere.
Remark 4.3. Even though, we have defined the Riemann curvature tensor from theLevi-Civita connection, every connection defines a curvature tensor.
Since R is a tensor field, for every p ∈ M , it defines a map defined on (Tp M)3,cf Section 2.14.
Given x, y ∈ Tp M , let Rx y be the map Tp M → Tp M sending z to Rx y z. We have
Proposition 4.1 (Symmetries of the Riemann tensor in the metric case ).
Rx y = −Ry x , (12)
g (Rx y v, w) = −g (Rx y w, v), (13)
Rx y z +Ry z x +Rzx y = 0, (14)
g (Rx y v, w) = g (Rv w x, y). (15)
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Remark 4.4. The curvature can be defined for any linear connexion. Some symme-tries always hold (such as the first), but others (such as the last) are specific to metricconnexion.
Remark 4.5. Equation (14) is often called the first Bianchi identity.
Proof. The first equation is obvious in view of the definition of R. For the second, weassume that X ,Y , Z are local vector fields such that Xp = x, Yp = y , Zp = z and suchthat all commutators [X ,Y ] = [Y , Z ] = [Z , X ] = 0. Since R is a tensor field, it does notdepend on the choice of extensions. (ex: why do such extensions exists ? ) Then,
g (RX Y V ,V ) = g (DX DY V ,V )− g (DY DX V ,V )
= X g (DY V ,V )− g (DY V ,DX V )−Y g (DX V ,V )+ g (DX V ,DY V )
= 1/2X Y g (V ,V )−1/2Y X g (V ,V )
= 0,
since X and Y commutes. By polarization, we obtain the second equation.We now consider equation (14). For a function F := F (X ,Y , Z ), let C (F ) denotes
the sum over cyclic permutations of X ,Y , Z i.e
C (F )(X ,Y , Z ) = F (X ,Y , Z )+F (Z , X ,Y )+F (Y , Z , X ).
Then, since any cyclic permutation leaves C (F )(X ,Y , Z ) unchanged, we have
C RX Y Z = C DX DY Z −C DY DX Z
= C DY DZ X −C DY DX Z
= C DY [Z , X ] = 0.
Equation (15) follows from the first three and we leave the computations as an exer-cice detailed below.
Exercise 4.1. With the above notation, consider the E(Y ,V , X ,W ) := g (C RY V X ,W ).Let H(Y ,V , X ,W ) be the sum over all cyclic permutationsσ(Y ,V , X ,W ) of (Y ,V , X ,W )of E(σ(Y ,V , X ,W )). Compute H as a sum of 12 terms by expanding the terms of theform C RY V X . Deduce that equation (15) holds.
Since R is a tensor field, we can consider its covariant differential DR : V ∈ Γ(M) →DV R. Recall that since DV is a tensor derivation, it maps tensor fields into tensorfields, and since DV R is F (M) linear in V , one can thus view DR as a (1,4) tensorfield which takes 4 vector fields X ,Y , Z ,V and return a vector field
DR(X ,Y , Z ,V ) = DV RX Y Z = (DV R)(X ,Y )V.
Again, at any p ∈ M , we get a map from (Tp M)4 to Tp M and given (x, y, z) ∈ (Tp M)3,Dz R(x, y) is an endomorphism of Tp M . We have
Proposition 4.2. We have
Dz R(x, y)+Dx R(y, z)+D y R(z, x) = 0,
called the second Bianchi identity.
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Proof. Choose a normal coordinate system (xα) at p. Let Xp ,Yp , Zp ∈ Tp M and letlet X ,Y , Z be local extensions of Xp ,Yp , Zp by choosing X , Y , Z to have constantcomponents in the ∂xα basis of vector fields. Note that X ,Y , Z all commutes togetherand that their covariant derivatives vanish at p.
Then, from the tensor derivation laws14,
(DZ R) (X ,Y )V = DZ (R(X ,Y )V )−R(DZ X ,Y )V −R(X ,DZ Y )V −R(X ,Y )(DZ V ).
At p, the two middle terms on the RHS vanishes, hence
(DZ R) (X ,Y )V = DZ [(DX DY −DY DX )V ]− (DX DY −DY DX )DZ V , at p (16)
= [DZ , [DY ,DX ]]V , at p,
where [DZ , [DY ,DX ]]V is a short-hand notation for the RHS of (16). The result thenfollows from the Jacobi identity
Let the components of the curvature tensor be defined as
R(X ,Y )Z = RαβγδZβX γY δ∂xα ,
i.e.Rα
βγδ∂xα = R(∂γ,∂δ)(∂β).
Exercise 4.2. 1. Show that
Rαβγδ = ∂γΓαδβ−∂δΓαγβ+ΓσδβΓαγσ−ΓσγβΓαδσ. (17)
2. LetRαβδγ = gαρRα
βγδ .
Show that Rαβγδ = Rγδαβ.
3. The Ricci identity: for X ∈ Γ(M), define the second covariant of X , DD X as thecovariant derivative D(D X ) of D X . It is a (1,2) tensor field and we denote itscomponents by Xα
;βγ. Prove that
Xα;βγ−Xα
;γβ = RαδβγX δ.
4. Let T be an arbitrary tensor field. Show that
[Dα,Dβ]T = (R?T )αβ.
where R?T denotes a tensor obtained from R ⊗T and contractions.
5. There are plenty of nice geometric interpretations of the Riemann curvature ten-sor. Write something about it. (look for instance for the relation between curva-ture and holonomy).
14Note that strictly speaking, we should apply DZ to the tensor R : (ω, X ,Y ,V ) → ω (R(X ,Y )V ). Onecan view the formula given above for DZ R(X ,Y )V as the correct definition such that DZ R (ω, X ,Y ,V ) =ω
(DZ R(X ,Y )V
).
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4.2 The Ricci tensor and the scalar curvature
Recall that a p-form on manifold is by definition a (0, p) tensor field which is totallyanti-symmetric. Given a semi-Riemannian manifold, we can define locally15 a n-form called volume form out of the metric, given by
η=√|det g |d x1 ∧d x2 ∧ ..∧d xn ,
where det g = det gαβ is the determinant of the metric in the local coordinate systemand d x1 ∧d x2..∧d xn is the unique n-form such that[
d x1 ∧d x2 ∧ ..∧d xn](∂x1 ,∂x2 , ...∂xn
)= 1.
Exercise 4.3. • Prove that if ω is a n-form and Vi = Ai j W j , where Vi , W j arevector fields and the Ai j are smooth functions, then
ω(V1, ..,Vn) = (det A)ω(W1, ..,Wn).
• Let (xα) and (y i ) be two local coordinate systems defined on a common open set.Let η be defined with respect to the (xα) coordinate system. Let det g y denote thedeterminant of g in the y coordinate system. Prove that
η=±√
|det g y |d y1 ∧ ..∧d yn .
• Prove that Dη= 0. (Hint: use normal coordinates.)
As we will see later, the volume form appears naturally in many formulae in-volving geometric operators and of course, it is also the natural n-form to performintegrations on our manifold.
Choose normal coordinates (xi ) at some p and consider the function det g =±[η(∂x1 ,∂x2 , ..,∂xn )
]2 in the (xα) system of coordinates. Doing a taylor expansion,we have
det g = det g (0)+∂α(det g )(0)xα+ 1
2∂α∂β(det g )(0)xαxβ+O(x3),
where det g (0) = ±1. The derivative terms on the right hand side can all be rewrit-ten in terms of covariant derivatives. Using the Levi-Civita property of the met-ric (Dg = 0), the symmetrization obtained thanks to the xαxβ, and the total anti-symmetry of the determinant and the special properties of normal coordinates, allthe second covariant derivatives can be rewritten using the Riemann curvature ten-sor. However, only specific combinations of the components of the curvature tensoractually appears and they can be rewritten as a tensor: the Ricci tensor, which isdefined as follows.
Definition 4.1. The Ricci tensor, denoted Ri c(g ), is the tensor obtained by contractionof the first and third positions of the Riemann curvature tensor, i.e. in coordinates
Ri c(g )βδ = Rαβαδ .
15It can even be made global if (and only if) M is orientable. See Appendix A.
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Remark 4.6. The Taylor expansion of the volume form in normal coordinates is givenby the formula √
|det g | = 1− 1
6Ri c(g )αβxαxβ+O(x3).
The Ricci tensor “measures” the change in volume in normal coordinates of an in-finitesimally small volume element. We include a proof of this formula in AppendixD below.
Lemma 4.2. The Ricci tensor is symmetric: Ri c(g )αβ = Ri c(g )βα.
Proof. This follows easily from the symmetries of the Riemann tensor.
Similarly, the scalar curvature is defined as
Definition 4.2. The scalar curvature is then defined as the trace of Ri c(g ) i.e. , in localcoordinates,
R(g ) = Ri cαβgαβ.
Remark 4.7. The scalar curvature has again an interpretation in terms of taylor ex-pansion, this time, one needs to consider the surface of a geodesic ball centered atsome point and compare it with the surface of a ball of flat space of the same radius.
5 Divergence free vector and tensor fields
5.1 Some differential operators
Definition 5.1. Let f ∈ F (M). Then, g r ad f (the gradient of f ) is by definition theunique vector field such that
g(g r ad f ,V
)=< d f ,V >=V ( f ),
for any V ∈ Γ(M).
In other words, g r ad f =] d f . In components,
(g r ad f )α = gαβ∂β f .
One often write D f instead of g r ad f for the gradient of f , which should thennot be confused the map D f : V → DV ( f ) =< d f ,V >= d f (V ).
Definition 5.2. Let X be a vector field. We then define its divergence di v(X ) as
di v(X ) := DαXα =C (D X )
where C is the contraction operator.
Exercise 5.1. 1. For a tensor field, we define similarly its divergence by computingfirst its covariant differential and then applying the contraction operator on thethe first two indices. For instance, the divergence of a (1,1) tensor field T is a oneform given in components by DµT µ
ν. Compute its components in terms of those
of T µν.
41
2. (a) Let (ai k ) be a non-singular matrix, of inverse ai k and determinant det a.Let (Ai k ) be the matrix of cofactors of (ai k ). We regard Ai k as smooth func-tions of the ai k . Prove that
∂ai j det a = Ai k = (det a)aki .
(b) Deduce from the first question that for a smooth metric g
∂xα det g = (det g )gγβ∂xαgγβ.
(c) Show that di v(X ) = 1|det g |1/2 ∂α
(|det g |1/2Xα)
in any system of coordinates
(note the absolute value, which is necessary in the Lorentzian case).
3. Let η = √|det g |d x1 ∧d x2..∧d xn be a local volume form associated to g andX ∈ Γ(M). Show that
d(iX η) = div(X )η=LX (η),
where iX η is an (n −1) form, the interior product of η by X defined by
(Hint: use a coordinate system such that ∂x1 = X .)
We end this section by the definition of a natural second order differential oper-ator associated to a Riemannian or Lorentzian metric.
Definition 5.3. • Assume that g is Riemannian, then the operator
ψ ∈F (M) →∆g (ψ) = gαβDαDβ = di v(g r ad f ),
is called the Laplace-Beltrami operator associated to g .
• For g Lorentzian, we have similarly an operator äg
ψ ∈F (M) →äg (ψ) = gαβDαDβ = di v(g r ad f ),
called the wave (or D’Alembertian) operator associated to g .
Remark 5.1. One can define a similar operator for any semi-riemannian metric g butit is only for the Riemannian and Lorentzian case that there currently exists a goodunderstanding (and, to the author knowledge, only these operators appears naturallyin physical contexts).
Remark 5.2. For a tensor field T of arbitrary type, we can define similarly
äg T = gαβDαDβT.
Remark 5.3. For a Riemannian metric, the principal symbol of ∆g written in an ar-bitrary system of local coordinates is elliptic, while for a Lorentzian metric it is hyper-bolic.
Exercise 5.2. • Show that the principal symbol of äg is gαβ(p)ξαξβ. It can thusbe viewed as a real function defined on the cotangent bundle T M?.
• Show that in any local system of coordinates such that g00 < 0, gi j definite pos-itive, if ξ = (τ,ξi ) (i.e. ξ0 = τ), then, for each (ξi ) 6= 0, the equation P (τ) :=gαβξαξβ = 0 has two distinct real roots.
• Show that in any local system of coordinates such that g00 < 0, gi j definite pos-itive, the wave equation can be recast as a symmetric hyperbolic system in thesense of Friedrichs.
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5.2 Remarks on conservation laws and Lagrangian theory
Recall Stokes theorem. Let M be a smooth orientable manifold and N ⊂ M a domainwith smooth boundary ∂N , j : ∂N → M the inclusion map, then ∀ω ∈Λn−1(M) suchthat supp (ω)∩∂N is compact, we have∫
∂Nj?ω=
∫N
dω.
Assume for simplicity that N is such that ∂N is compact and let X be a vectorfield.Assume also that M is orientable. In this case, the volume form η can actually bedefined globally and agree with the given formula in any local chart16. Then, fromExercise 5.1.2c, ∫
∂Nj?(iX η) =
∫N
div(X )η.
In particular, for every divergence free vector field X ,∫∂N
j?(iX η) = 0.
In physics, this is interpreted as a conservation law, in the sense that the totalflux of X through ∂N vanishes, so that the incoming flux=the outgoing flux.
These conservation laws are important in PDEs (and of course in physics) appli-cations because they will be the source of many of the estimates needed to controlsolutions.
The question becomes: how to construct divergence free vector fields ?One answer: out of divergence free, symmetric tensor fields and out of symmetries
of the metric.To explain this, let T be a divergence free tensor field. Given T µ
ν its components,the divergence free condition reads
DµT µν = 0.
Let now X be a vector field and define the vector field X J by its components as
JX µ = T µν X ν.
Then,div( JX µ) = T µ
ν DµX ν.
Thus, JX is divergence free and we get a conservation law provided that T µν DµX ν =
0.Assume now that T is a symmetric tensor field, so that T µν = T νµ. In this case,
T µν DµX ν = T µν1/2
(DµXν+DνXµ
)= T µνXπµν,
where πX is the deformation tensor of X .Recall from Definitions 3.11-3.12 and Lemma 3.8 that vector fields with vanish-
ing deformation tensor, called Killing fields, arose from the symmetries of the space-time (their flow generates isometries).
Conclusion: Killing vector fields and symmetric divergence free tensors give con-servation laws as explained above.
16See Appendix A
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Remark 5.4. Similar arguments based on stokes theorem can be applied to symmetrictensor fields and vector fields even when they are not divergence free or Killing. In thatcase, we would get error terms from the divergence term, and the game becomes tocontrol these errors. In particular, the control of solutions for non-linear problemsoften falls in that category.
The question now becomes: how to construct or obtain divergence free symmet-ric tensors and Killing fields ? It turns out that in many physical problems, both arisenaturally.
5.3 The energy-momentum tensor
For all the standard equations of physics (say coming from electromagnetism, fluiddynamics, scalar fields etc..), it is not hard to write them down first in tensorial no-tation in special relativity (i.e. in the special case of a background geometry given byMinkowski space) and then in an arbitrary Lorentzian manifold by replacing partialderivatives in Cartesian coordinates by covariant derivatives in local coordinates.Now, in all these cases, the equation can be written simply under the form of a sym-metric divergence free tensor field constructed out of the unknowns. These diver-gence free tensor fields themselves can be constructed from variational techniquesand takes their roots in Lagrangian theory17. Thus, the physics is essentially goingto tell us what are the divergence free symmetric tensor to consider. As for Killingfields, they will appear also naturally. For instance, if we imagine a physical systemwhich is stationary (i.e. there is a sense in which the system is not evolving), thenthe physicist will model this by the existence of a Killing field T which has sometimelike property (g (T,T ) < 0 at least in some region of the manifold). There is an-other (similarly vague) argument for the existence of symmetry in many solutions ofproblems arising in physics. These solutions can often be constructed by variationalarguments, for instance by constructing minimizers of certain energy functionals.Departure of symmetry will then often enhance the value of these functionals. Agood example to have in mind is the isoperimetric problem in classical geometry.
We now provide two classical examples of equations and their associated diver-gence free tensor field.
Example 1 We consider a smooth function ψ : M →R (a scalar field) on our manifold andassume that it solves the wave equation
ägψ= F, (18)
for some smooth source F : M →R.
To any suchψ, we can associate its so called energy-momentum18 tensor (field)
T [ψ] = dψ⊗dψ− 1
2g
(g (Dψ,Dψ)
).
Then,
Lemma 5.1.divT =ägψ.dψ,
or in coordinates,DµT µ
ν [ψ] =ägψ.Dνψ.17See for instance [Chr00] as well as the easier presentation given in [Chr08].18Also refers to sometimes as stress energy tensor.
44
In particular, T is divergence free provided thatψ is a solution of the homoge-neous wave equation, i.e. F = 0 in (18).
Example 2: Maxwell equations. In Minkowski space in usual coordinates, they can bewritten19 as
∂µFµν = jν, (19)
where jν is a divergence free vector field (the source term, sometimes calleda 4-current) and F (the Faraday tensor) is a closed two form on Rn+1 (thus, bythe Poincaré Lemma in Rn , F = d A for some one-form A).
Exercise 5.3. Derive from (19) the usual form of the Maxwell equations (in-volving the divergence and rotational operators in flat space) in terms of theelectric and magnetic fields Ei = F (∂x0 ,∂xi ) and Bi =− 1
2εi abδa jδbk F (∂x j ,∂xk ),
where ε is the totally antisymmetric symbol such that ε123 = 1.
We then introduce the energy-momentum tensor of electromagnetism givenin components by
Tµν[F ] = FµσFσν−
1
4ηµν(FαβFαβ).
Exercise 5.4. Check that the equation of motion (19) implies that
∂µTµν = Fµν jν.
In particular, in the vacuum j = 0 and we again have a divergence free sym-metric tensor field.
To obtain the equations in an arbitrary Lorentzian manifold (M , g ), just re-place all ∂µ by Dµ and the Minkowski metric η by g . Thus, the Maxwell equa-tions in the vacuum (no source term) for a closed two form F is simply givenby
DµFµν = 0.
We end this section with another useful property of the energy-momentum ten-sors.
Proposition 5.1. Let X1, X2 be two timelike vectors such that g (X1, X2) ≤ 0. Let T [ψ]and T [F ] be the energy-momentum tensor fields as defined above associated to a ei-ther a smooth function or a 2-form. Then,
T [ψ](X1, X2) ≥ 0, T [F ](X1, X2) ≥ 0.
In physics, these positivity properties are interpreted as positivity of local energydensities. For pde applications, these positivity properties allow to get good esti-mates from the conservation laws. The positivity property also extends to causalvectors.
Exercise 5.5. 1. Prove the proposition. What about the equality cases ?
2. We consider again Stokes theorem in a domain N . We assume that T is theenergy-momentum tensor of a scalarψ of Maxwell field F and that the supportof ψ or F is contained in a region where ∂N is spacelike and orientable.
• Prove that the normal to ∂N is a timelike vector field on ∂N .
• Deduce that on each connected component of ∂N , T [ψ](X1, X1) has a sign.19As for the rest of the equations written in these lectures, all physical constants have been set to 1 for
simplicity.
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5.4 The Einstein tensor
Let us now try to construct general relativity. In particular, we are looking for anequation replacing Poisson’s equation
∆φ= 4πρ.
Let T be the energy-momentum tensor associated with the matter fields as intro-duced in the previous section. Recall that in the above examples, T is (0,2) symmet-ric tensor field and T is divergence free. Moreover, from Proposition 5.1, the energy-momentum tensor has some positivity property, which is analogous to ρ ≥ 0.
We interpret T as the correct tensorial replacement for ρ. On the other hand,the replacement for φ should be constructed out of the metric. Moreover, it shouldbe second order (as is ∆φ) in terms of derivatives of g . Thus, we want an equationunder the form
G(g ) = T,
where G is a symmetric (0,2) tensor (it has to be independent of the choice of coor-dinates), constructed out of the first and second derivatives of g and G is divergencefree. A first candidate would have been the curvature tensor, since it is a tensor anddepends on the second derivatives of g , but it is 4 tensor. The Ricci tensor thenseems like a good candidate and indeed, in the first draft, Einstein (wrongly) choseit. The trouble is that Ri cci is not divergence free in general.
The correct choice is given in the next lemma.
Lemma 5.2. The differential dS(g ) of the scalar curvature S(g ) verifies
dS(g ) = 2divRi c.
Equivalenty, the Einstein tensor defined as
G = Ri c(g )− 1
2g S(g )
is a divergence free (symmetric) tensor field.
Proof. Note that since Ri c(g ) is a symmetric (0,2) tensor field, by definition, divRi cis the 1-form of components
divRi cγ := gαβDβRi cαγ
= DαRi cαγ
= DαRi cαγ= Ri cαγ;α .
We first express the second Bianchi identity in coordinates
Rαβγδ;ρ +Rα
βργ;δ+Rαβδρ;γ = 0
Contracting in α and ρ gives, using also the symmetries of R and the definition ofthe Ricci tensor, ‡
0 = Rαβγδ;α+Rα
βαγ;δ+Rαβδα;γ
= Rαβγδ;α+Ri cβγ;δ−Ri cβδ;γ.
Contract again in βδ (and use again the symmetries of R) to conclude.
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6 The Einstein equations
The Einstein equations for a Lorentzian20 manifold (M , g ) are given by
Ri c(g )− 1
2g S(g )+Λg = T,
whereΛ ∈R is a constant, typically called in the physics litterature the cosmologicalconstant, T is a symmetric, divergence free, 2-tensor, the energy-momentum tensor(think of T as a source term depending on the choices of matter models). Note thatthe termΛg is also divergence free, by virtue of the fact that Dg = 0.
For simplicity, let us assume that T = 0 and that dim M = 4. Then, taking thetrace of the Einstein equations, we obtain that S(g ) = 4Λ and the Einstein equationsreduce to the vacuum Einstein equation
Ri c(g ) =Λg .
A solution to the Einstein vacuum equations is then a Lorentzian manifold (M , g )solving the above equations. Some remarks:
1. The manifold M is part of the unknown. We are not just solving for g . This issimilar to ode theory, where a solution to an ode is a couple ( f , I ), where I isan interval and f a solution to the ode defined on I .
2. A word about the real constant Λ. In the first formulation of the equations,Einstein did not add the Λ term. Then, trying to construct explicit static so-lutions (which have a time symmetry ) in the presence of matter, he neededto balance the gravitational force that attracts matter together with anothermechanism that stretch matter. Adding Λ> 0 allowed to construct such solu-tions. The solution that Einstein constructed is however highly unstable, thusnot quite physical and moreover does not match the observations that Hub-ble made in 1929. He called the introduction of theΛ is "biggest mistake" anddecided to remove it from the equations. However, even if the solution of Ein-stein did not match the physical observations, the new mechanism comingfrom Λ > 0 was leggit. It is responsible for the so-called "accelerated expan-sion" of the universe which nowdays a standard of cosmology. In other words,yes the solution of Einstein was unphysical but that does not necessarily meanthe equations were bad. There could be other solutions with better properties.The current models in astrophysics and cosmology use Λ = 0 for an event atthe scale of a galaxy or a star andΛ> 0 (but very small21 compared to all otherconstants, such as G and c)) for a global model of the observed universe. Fi-nally, solutions with Λ < 0 are extremely popular in High Energy Physics (themost quoted paper in HEP is about such type of solutions).
3. As we shall see, the Einstein equations are dynamical (hyperbolic) equations.Thus, solutions are typically constructed by solving initial value problem (butother constructions are possible such as characteristic initial value problem,analytic extensions...) in which case we will need to explain what are the dataand in which sense our solutions agree with the data.
20The study of the Riemannian version of these equations is also an interesting subject. Here, we willfocus on the Lorentzian case.
21According to the English wikipedia article the current proposed value is 1.19×10−52m−2.
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4. In the non-vacuum case (T 6= 0), even though T is a source, the typical caseis that T is not given a priori. Instead, T is also obtained by solving a dynam-ical equation, which itself depends on (M , g ). Thus, one obtained a coupledsystem.
Example: The Einstein-scalar field system. Consider a functionψ ∈F (M) andrecall that T [ψ] is given by
T [ψ] = dψ⊗dψ− 1
2g (g (Dψ,Dψ),
where Dψ denotes the gradient of ψ. Recall also that
divT = Dψ .ägψ.
Thus, we consider the coupled problem
Ri c(g )− 1
2g +Λg = T [ψ],
ägψ = 0.
Typical other sources term: electromagnetism, Vlasov fields (kinetic theory),fluids etc..
Above, we claimed that the Einstein equations are actually hyperbolic, in factwave, equations. There are several different notions of hyperbolicity in the PDE lit-erature. Here, we are going to prove first that, from the Einstein equations, one canextract systems of wave equations, that is to say equations of the form
äg u = F (u,Du)+B ,
for some tensor fields u related to the metric (These equations may only hold locally,say in some coordinate patch of the manifold).
6.1 A model problem: the Maxwell equations
The difficulties in identifying the hyperbolic character of the Einstein equations areactually already present in a much simpler, and better known systems of PDEs: theMaxwell equations.
Recall that they can be written (here in Minkowski space for simplicity) as ∂µFµν =jν for a closed two form F .
How do we know that these are actually wave equations ?We will give two answers22.First, in the (E ,B) formulation, it is well known that from the above system we
can write wave equations of the form
äE = SE (∂ j ), äB = SB (∂ j ).
22A third formulation of hyperbolicity for the Maxwell equations can be obtained directly from the(E ,B) formulation. Discarding the two divergence equations, the two evolution equations for E and Bactually form a 1st order symmetric hyperbolic system in the sense of Friedrich. Interestingly, the secondBianchi equations for the curvature tensor can also be separated into constraints and evolution equationsand the evolution part forms again a symmetric hyperbolic system. That there are so many analogies be-tween the maxwell equations DµFµν and the Bianchi equations is actually not that surprising, since theMaxwell equations can also be viewed as the Bianchi equations for a connection 1-form on a principalbundle whose curvature is actually given by F .
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Second, if A is such that F = d A, then the Maxwell equations written in terms ofA takes the form
äAν−∂ν∂µAµ = jµ.
Now this equation is not a wave equation because of the second term in the left-hand side. However, recall that given any scalar function χ, we have F = d(A +dχ).Claim: if χ solves äχ=−∂µAµ, then A′ = A+dχ solves ∂µ(A′)µ = 0. The new A′ thensolves a wave equation. Choosing χ is called making gauge choice (the one where∂µAµ = 0 is called the Lorentz gauge, another popular one is the Coulomb gauge,where only the spatial divergence of A is required to vanish ∂i Ai = 0) .
For the Einstein equations, we claim that similarly, we can write wave equationsdirectly by essentially differentiating the equations (thus obtaining wave equationsfor R) or, that by making a gauge choice, we can write wave equations for g . Thegauge freedom in this case is simply the choice of local coordinates on our manifold.
6.2 Hyperbolicty I: Wave equations satisfied by the Riemann tensor
Proposition 6.1. Let (M , g ) be a solution to the vacuum Einstein equations Ri c(g ) =Λg . Then, the Riemann curvature tensor solves an equation of the form
äg R = R?R.
where R?R is obtained from R ⊗R by contractions.
Proof. Taking the trace of the Bianchi equations, we obtain
DσRαβσν+DβRαν−DαRβν = 0.
Thanks to the Einstein equations and the fact that Dg = 0, by definition of the Levi-Civita connection, the last two terms vanishes and we obtain that the Riemann cur-vature tensor is divergence free
Remark 6.1. Given (M , g ) and apropriate data, it is not hard to construct local solu-tions to a non-linear system equations of the form
ägφ=φ?φ.
However, this scheme does not really solves the Einstein equations in that R and g areof course not independent. We could imagine running a scheme of the form
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1. Given appropriate data for g , we construct first a candidate for g 1 (for instanceextending g locally to simply be a Lorentzian metric, whatever it is, as long as itagrees with the data).
2. Then, we compute data for R1 and solve the associated wave equation.
3. We then try to construct g 2, which would have R1 as a curvature tensor (possiblymodulo some error terms) and should still agree with the data.
4. We then run again our iterative scheme and hope to prove convergence.
There are many difficulties in doing so and in practice, the standard (and "fastest")proof of local existence of solutions to the Einstein equations does not use this formu-lation.
However, estimating the curvature first and returning to g has been a very success-full strategy to address global problems for the Einstein equations, where one tries tounderstand the long time dynamics of the solutions. The important point which isexploited there is that the wave equation satisified by R is independent of anything: itis gauge free.
6.3 Hyperbolicity II: Wave coordinates
Definition 6.1. Let (M , g ) be a Lorentzian manifold. A system of wave coordinates23
is by definition a coordinate system (xα) such that for each α, the function xα is itselfa solution of the wave equation
äg xα = 0.
Proposition 6.2. Let (xα) be wave coordinates. Then, for all α,
gρσΓαρσ = 0.
Proof. We have for any function ψ
ägψ = gρσDρ∂σ(ψ)
= gρσ(∂ρ∂σ(ψ)−Γγρσ∂γ(ψ)
).
Thus,
äg xα = gρσ(Γγρσδ
αγ
)= gρσΓαρσ.
Proposition 6.3. In wave coordinates, the wave operator äg reduces to its principalpart and the vacuum Einstein equations Ri c(g ) =Λg takes the form
äg gαβ =Qαβ(∂g ,∂g )−2Λg ,
where Qαβ(∂g ,∂g ) is quadratic in the first derivatives of g .
23In the Riemannian case, similar coordinates can be constructed, replacing the wave operator by theLaplace-Beltrami operator. These are called harmonic coordinates and this name is sometimes also usedin the Lorentzian case.
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Proof. Recall from equation (17), that the components of the curvature tensor aregiven by
Rαβγδ = ∂γΓαδβ−∂δΓαγβ+ΓσδβΓαγσ−ΓσγβΓαδσ,
so that
Ri cβδ = ∂αΓαδβ−∂δΓααβ+Qαβ(∂g ,∂g ), (20)
where, since we are interested only in the terms containing second order derivatives,we denote by a generic Q any terms quadratic in ∂g (and not dependent on ∂2g ).Below, we continue to write any such terms by Q but it may change from line to line.
Define Γαβγ as
Γαβγ := 1
2
(gγβ,α+ gαβ,γ− gαγ,β
),
so thatΓµ
αβ= gµγΓαγβ.
With this notation,
∂νΓνµρ = ∂ν
(gνβ
)Γµβρ + gνβ∂νΓµβρ ,
∂µΓννρ = ∂µ
(gνβ
)Γνβρ + gνβ∂µΓνβρ .
Thus, from (20), in order to compute Ri cµρ , we need to compute
gνβ∂νΓµβρ − gνβ∂µΓνβρ .
Moreover,
2∂νΓµβρ = gβρ,µν+ gµβ,ρν− gµρ,βν,
2∂µΓνβρ = gβρ,νµ+ gνβ,ρµ− gνρ,βµ.
so that
2gνβ(∂νΓµβρ −∂µΓνβρ
) = −gνβgµρ,βν+ gνβ(gµβ,ρν− gνβ,ρµ+ gνρ,βµ
).
The first term on the RHS is −äg gµρ , since from the wave coordinate conditions,äg reduce to its principal part. Thus, we would like the second term on the RHS tobe a Q term. This means that the second order derivatives terms must cancel. Forthese to happen, we need to use again the wave coordinate condition. In terms ofthe Γβµν, the wave coordinate conditions reads
gµβΓβµν = 0.
Thus, we try to rewrite the second terms in terms of the Γ.For this, since gνβ is symmetric in (νβ), we rewrite the second term on the RHS
using the wave coordinate condition.Thus, in wave coordinates, we have
Ri c(g )αβ =−1
2äg gαβ+Qαβ,
so that the vacuum Einstein equations imply that
äg g =−2Λg +Q.
Exercise 6.1. The lower order terms play no role as far as understanding the wavecharacter of the Einstein equations but they are important for more precise state-ments, for instance, to describe the global behaviour of solutions.
Prove that, in wave coordinates,
Rµρ =−1
2gαβ∂αβgµρ + gαβgγδ
(ΓαγµΓβδρ +ΓαγµΓβρδ+ΓαγρΓβµδ
).
Remark 6.2. The existence of local wave coordinates will be relatively easy once weknow how to solve the wave equation.
Remark 6.3. We have just proved that the vacuum Einstein equations take the form
gαβ∂xα∂xβg =Q(∂g ,∂g )+CΛg (21)
in wave coordinates. The equations (21) form a standard system of quasilinear waveequations, which can be solved locally by standard methods, with initial data g ,∂x0 ggiven on some x0 = const slice provided for instance that g 00 < 0 and (g i j ) > 0 is defi-nite positive initially. See for instance [Sog95a, Section I.4]. On the other hand, to get alocal solution to the full Einstein equations, one then needs to verify, after solving theequations, that the coordinates are then wave coordinates for the newly constructedmetric. Moreover, one needs to explain where the data for g is coming from and ex-plain what do we mean by a x0 = const slice while we have not yet constructed M!Finally, in general, one needs several system of coordinates to cover our manifolds andtherefore, one obtains several local solutions. One thus needs some way to patch allthese solutions together to get a complete development of the data.
Remark 6.4. For more on the gauge invariance of the Einstein or Maxwell equationsand its relation to the hyperbolicity of the equations, one can consult [Chr08, Chapter2.1.1].
7 Semi-Riemannian submanifolds
Recall the definition of a vector field on a map, Definition 3.13 as well as the defi-nition of a submanifold Definition 2.10. If P is a submanifold of M , then we call avector field on the inclusion map j : P → M a P-vector field. We denote by Γ(P, M)the set of all such vector fields. Note that Γ(P ), the set of vector fields of P can beviewed as a subset of Γ(P, M).
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7.1 The induced connection
Definition 7.1. Let P be a submanifold of M. Then the induced connection ∇ on Pis a map
∇ : Γ(P )×Γ(P, M) → Γ(P, M)
defined as follows. For any V ∈ Γ(P ) and Z ∈ Γ(P, M), let V ′ and Z ′ be smooth localextensions to M of V and Z . Then we define ∇V Z as the restriction of DV ′ Z ′ to P.
Exercise 7.1. 1. With the above notation, check that the restriction does not de-pend on the choice of extensions, and thus that the induced connection is welldefined.
2. Redefine the induced connection using the pull-back bundle j∗(T M).
We have immediately from the definition.
Lemma 7.1. Let ∇ be the induced connection on P. Then, if V ,W ∈ Γ(P ) and X ,Y ∈Γ(P, M), we have
1. ∇V X is F (M)-linear in V and R linear in X
2.∇V ( f X ) =V ( f )X + f ∇V X ,
3.[V ,W ] =∇V W −∇W V ,
4.V g (X ,Y ) = g (∇V X ,Y )+ g (X ,∇V Y ).
7.2 Semi-Riemannian submanifold ans the Levi-Civita connectionof the induced metric
The above definition make sense for any submanifold of M . Recall that in general,the pullback of the metric on a submanifold is not necessarily a semi-Riemannianmanifold cf 3.4.
Definition 7.2. Let P be a submanifold of a semi-Riemannian manifold (M , g ) andj : P → M denote the inclusion map. If j?(g ) is a metric tensor on P, we say that P issemi-Riemannian submanifold of M.
j?(g ) is called the induced metric or first fundamental form.A submanifold of a Riemannian or Lorentzian manifold is called spacelike if the
induced metric is Riemannian, timelike if the induced metric is Lorentzian, null ifthe induced metric24 is degenerate. In the case of a submanifold of codimension 1,we say that we have a spacelike, null, or timelike hypersurface.
If P is a semi-Riemannian submanifold of M , then we have, for each p ∈ M adirect sum decomposition
Tp M = Tp P + (Tp P )⊥,
where(Tp P )⊥ = v ∈ Tp M : g (v, x) = 0, ∀x ∈ P .
24This is of course a slight abuse of language, since the induced metric in that case is not a metric.
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Vectors in (Tp P )⊥ are said to be normal to P while those of Tp P are of coursetangent to P . The resulting orthogonal projection will be denoted as
tan : Tp M → Tp P
andnor : Tp M → (Tp P )⊥.
A vector field Z in Γ(P, M) is normal (respectively tangent) to M provided that foreach p ∈ P , Zp is normal (respectively tangent) to P . We denote the set of all normalvector fieds by Γ(P )⊥, called the normale bundle.
For any Z in Γ(P, M), we can apply tan and nor at each point to construct thetangential and normal part of Z .
In general, if V ,W are both tangent to P , it does not follow that ∇V W is tangentto P . In the case of a semi-Riemannian submanifold, we have however
Lemma 7.2. If V ,W ∈ Γ(P ) and P D denotes the Levi-Civita connection associated tothe induced metric on P, then
tan∇V W = P DV W.
Remark 7.1. In particular, we could reconstruct the properties of the Levi-Civita con-nection for surfaces (or any submanifolds of Rn) in this way. Indeed, let S be a sur-face in R3. Then, we know how to define the induced metric on S. Moreover, forV ,W ∈ Γ(P ), we can consider extensions V , W to the whole of Rn . Then, ∇V W is givenin Cartesian coordinates by V α(W β) and we can restrict to the surface as explainedabove. In other words, the formula just given together with basic calculus in Rn canbe seen as justification for the definition of the Levi-Civita connection.
Proof. Recall the Koszul formula for the Levi-Civita connection
2g (DV W, X ) =V g (W, X )+W g (X ,V )−X g (V ,W )−g (V , [W, X ])+g (W, [X ,V ])+g (X , [V ,W ]).
Suppose that V ,W, X are all vector fields on P and consider extensions V , W , X toM . We have
2g (DV W , X ) = V g (W , X )+W g (X ,V )−X g (V ,W )−g (V , [W , X ])+g (W , [X ,V ])+g (X , [V ,W ]).
Restricting to P , we obtain the equality
2g (∇V W, X ) = V g (W, X )+W g (X ,V )−X g (V ,W )− g (V , [W, X ])+ g (W, [X ,V ])+ g (X , [V ,W ])
= V h(W, X )+W h(X ,V )−X h(V ,W )−h(V , [W, X ])+h(W, [X ,V ])+h(X , [V ,W ]),
where h is the induced metric on P . Since X is tangent to P , we have
2g (∇V W, X ) = 2g (tan∇V W, X ).
It follows that the map(V ,W ) → tan∇V W
1. defines a connection, 2. satisfies the Koszul formula for the induced metric. Byuniqueness, it must agree with the Levi-Civita connection of h.
In the rest of this section, P will denote a semi-Riemannian submanifold of M .
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7.3 The shape tensor
For the normal part, we have
Lemma 7.3. The function Π : Γ(P )×Γ(P ) → Γ(P )⊥ given by
Π(V ,W ) = nor∇V W
defines a symmetric tensor (in the sense of tensor bundles), i.e. it is symmetric andF (P )-bilinear.
Proof. The tensorial part follows easily from the definition. For the symmetric part,we have
Π(V ,W )−Π(W,V ) = nor(∇V W −∇W V ) (22)
= nor[V ,W ] = 0. (23)
From the above, we have
∇V W = DPV W +Π(V ,W ).
Note that in the case of (non-characteristic) hypersurfaces, the normal bundle isone dimensional25 in the sense that every normal vector to P must be propotionalto a unit normal. It follows that we can write Π(V ,W ) = k(V ,W )N , where N is aunit normal and k is a (0,2)-tensor field. The tensor Π is called the shape tensor orsecond fundamental form, this last name being also used for the tensor k.
7.4 The Gauss equation
The relation between the curvature of P and that of M is given by the Gauss equation
Theorem 7.1. Let RP denote the Riemann curvature tensor of P endowed the inducedmetric h. Then, for V ,W, X ,Y ∈ Γ(P )
h(
RP (V ,W )X ,Y)= g (R(V ,W )X ,Y )+ g (Π(V , X ),Π(W,Y ))− g (Π(V ,Y ),Π(W, X )),
where R(V ,W )X denotes (by a small abuse of language) R(V ,W )X restricted to P,where V ,W , X are extensions to M of V ,W, X .
Proof. Since the equation is tensorial, we can assume wlog that [V ,W ] = 0. With asmall abuse of notations, we consider extensions of all vector fields to M and denotethem by the same letter.
We can write schematically
g (R(V ,W )X ,Y ) = (V W )− (W V ),
whereV W = g (DV DW X ,Y ).
25Strictly speaking it is a (n −1)+1 dimensional manifold.
55
Let Z = DW X . By definition Z is a vector field on M and its restriction to P is a vectorfield over P . By definition, this vector field is in fact ∇W X . Moreover, the restrictionof DV Z to P is also by definition ∇V Z . Thus, on P , we have
V W = g (∇V ∇W X ,Y )
= g (∇V(DP
W X +Π(W, X ))
,Y )
= g (tan∇V DPW X ,Y )+ g (∇VΠ(W, X ),Y )
= h(DPV DP
W X ,Y )+V g (Π(W, X ),Y )− g (Π(W, X ),∇V Y )
= h(DPV DP
W X ,Y )− g (Π(W, X ),Π(V ,Y )).
Computing the difference (V W )− (W V ), we get the result.
7.5 The normal connection and the Codazzi equation
In the previous section, we considered the geometry of vectors tangent to P . Here,we will consider the geometry of vectors normal to P .
Definition 7.3. The normal connection of P is the map
∇⊥ : Γ(P )×Γ(P )⊥ → Γ(P )⊥
defined by∇⊥
V Z = nor ∇V Z ,
where ∇V Z is the induced covariant derivative and Γ(P )⊥ is considered as a subset ofΓ(P, M).
∇⊥V Z is called the normal covariant derivative of Z with respect to V . The follow-
ing properties are immediate consequence of the definition.
Lemma 7.4. If V ∈ Γ(P ) and Y , Z ∈ Γ(P )⊥, then
1. ∇⊥V Z is F (M)-linear in V and R-linear in Z
2.∇⊥
V ( f Z ) =V ( f )Z + f ∇⊥V Z ,
3.V g (Y , Z ) = g (∇⊥
V Y , Z )+ g (Y ,∇⊥V Z ).
Since the shape tensor is a tensor valued in Γ(P )⊥, we cannot apply our standarddefinition of covariant derivative to it. Instead, we will use the following definition.
Definition 7.4. Let Π be the shape tensor of P and V , X ,Y three vector fields in Γ(P ).We define the covariant derivative ofΠwith respect to V by
DPV Π(X ,Y ) =∇⊥
V (Π(X ,Y ))−Π( DPV X ,Y )−Π(X , DP
V Y ).
AsΠ, DPV Π(X ,Y ) is a symmetric tensor with values in Γ(P )⊥.
Exercise 7.2. Let P be a semi-Riemannian hypersurface with unit normal N (N couldbe defined locally only if P is non-orientable) i.e. g (N , N ) =±1 = ε.
1. Prove thatΠ(X ,Y ) = k(X ,Y )N , with k(X ,Y ) = εg (N ,∇X Y ).
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2. Prove that for any vector field V ∈ Γ(P ), ∇V N ∈ Γ(P ).
3. Prove that ∇⊥(Π(X ,Y )) =V (k(X ,Y ))N .
4. Prove that ( DPV Π)(X ,Y ) = ( DP
V k)(X ,Y )N .
The Gauss equation describes tanR(V ,W )X , where V ,W, X where are all tangentto P in terms of the instrinsic curvature of P and its shape tensor. The Codazzi equa-tion completes this information by providing nor R(V ,W )X .
Proposition 7.1. For V ,W, X ∈ Γ(P ), we have
nor R(V ,W )X =−( DPV Π)(W, X )+ ( DP
V Π)(V , X ).
7.6 The constraint equations
From the Gauss-Codazzi equations, one obtain the following
Proposition 7.2. Let (M , g ) be a Lorentzian manifold and P a spacelike hypersurface.We assume that P admit a unit normal N . Then, on P, we have
G(N , N ) = 1
2SP −ki j k i j + (trhk)2, (24)
G(N , v) = DP j kki − DPi (trhk)v i . (25)
7.7 Normal frame field
Definition 7.5. Given a semi-Riemannian submanifold P of dimension r and U ⊂ Pan open submanifold of P a (local) normal frame field in U is an orthornormal familyof n − r vector fields in Γ(U )⊥.
Example:Consider a submanifold P in M and an adapted local coordinate sys-tem (xα) in which P is given locally by the equations x1 = 0, x2 = 0. The vector fields∂xk , k > 2 then span (locally) the tangent bundle of P . A vector field X is normal toP provided g (X ,∂xk ) = 0 for any k > 2. In coordinates, this reads as the conditions
Xαgαk = 0.
In another words, the one form X[ = Xαgαβd xβ is of the form X[ = a1d x1 + a2d x2
and X is of the form X = gα1a1∂xα + gα2a2∂xα . Let now
X1 = gα1a1∂xα + gα2a2∂xα
X2 = gα1b1∂xα + gα2b2∂xα .
(X1, X2) is orthonormal provided the equations
g 11a1b1 + g 22a2b2 + g 21a1b2 + g 12a2b1 = 0
g 11a21 + g 22a2
2 +2g 21a1a2 = ±g 11b2
1 + g 22b22 +2g 21b1b2 = ±
are satisfied. If for instance g 11 6= 0 and we are in the Riemannian case, then thesigns on the RHS are all + and we can take a1 = (g 11)−1/2, b2 = (g 11)1/2/
√det g , b1 =
−g 12/g 11b2 and obtain a normal frame field.
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More generally, consider a submanifold P and and an adapted local coordinatesystem (xα) in which P is given locally by the equations x1 = x2 = .. = xr = 0. Asabove, the one-forms which are normal to P are of the form
ω=n∑
i=r+1ai d xi .
Consider a non-zero one-form ω1 of the above form. Since P is assume to be non-characterisctic, we can choose ω1 to be a non-null one-form and then normalizedit. Similar to the Graham-Schmidt, we can then consider an independent one-formω2, such that ω2 is also non-null and normalized. Repeating we then obtain an or-thonormal family of one-forms and the corresponding family of vector fields thenform a normal frame field. In particular, every semi-Riemannian submanifold ad-mits local normal frame fields.
Exercise 7.3. Generalize the notion of normal frame field by considering null frames.
7.8 The normal bundle
Recall the definition of a vector bundle over M , definition 2.19. Let P be a semi-Riemannian manifold of M and let N P be the set of all points (p, vp ) where p ∈ Pand vp ∈ Tp P⊥. Let π : N P → P be the natural projection. For each p in P , thereexists a normal frame field (E1, ..,Ek ) defined in a neighborhood U of p in P . N P canbe given the structure of a manifold, for instance, we can use locally a coordinatesystem in an neighborhood V ⊂ U using a local coordinate system (V , xα) of thebase manifold P together with the coordinate systems induced by the Ei on eachTp P⊥.
Define a map φ : U ×Rk → N P by
φ(q, a1, .., ak ) =k∑
i=1ai [Ei ]q .
This defines a smooth map which is a diffeomorphism from U ×Rk to π−1(U ) andmakes N P a vector bundle over P .
The sections of N P are then the vector fields in Γ(P )⊥.
Definition 7.6. For any p in P let D⊥p be the set of all vectors v in Tp P⊥ such that
the geodesic starting at p with initial velocity v is well defined up to parameter 1. LetD⊥ =⋃
p∈P D⊥p . Then, we define the normal exponential map as
exp⊥ : D⊥p ⊂ N P → M
(p, v) → γv (1).
As for the exponential map, D is an open set containing all points (p,0p ) andexp⊥ is a smooth map. Let Z be the set of all points (p,0p ).
Definition 7.7. A neighborhood U of P is said to be normal if U is the image by exp⊥of a neighborhood of Z in N P.
Lemma 7.5. If p ∈ P, then exp⊥ carries some neighborhood of (p,0p ) in NP diffeo-morphically onto a neighboord of p in M.
58
Proof. Z is a submanifold of N P (of dimension that of P ). Thus, exp⊥ restricts toa smooth map from Z to P . This is in fact a diffeomorphism of inverse given bythe zero vector field. Thus, d exp⊥
(p,0p ) is one-to-one from T(p,0p )Z to Tp P . Consider
now d exp⊥(p,0p ) restricted to T(p,0p )Tp P⊥. As in the case of the exponential from a
point, this is the canonical isomorpshim T(p,0p )Tp P⊥ ' Tp P⊥. Since we have the
direct sum T(p,0p )D⊥p = T(p,0p )Z
⊕T(p,0p )Tp P⊥, it follows that d exp⊥
(p,0p ) is a linear
isomorphism and we can apply the inverse function theorem.
We then prove
Proposition 7.3. Ever semi-Riemannian submanifold P has a normal neighborhood.
Proof. For each p ∈ P , let Np be a neighborhood of (p,0p ) in N P on which exp isa diffeomorphism. By shrinking Np , we can arrange that if exp(v) ∈ P for v ∈ Np ,then v = 0. Moreover, we can assume that Np has compact closure. Let N =⋃
Np .One easily have that for any K ⊂ P compact, π−1 ∩ N is compact. Note that byconstruction, if v 6= w and exp(v) = exp(w), then v = w = 0.
Consider a decreasing sequence of neighborhood of Z of the form N = N0 ⊃N1 ⊂ ... such that
⋂i Ni = Z . For any K , there is an i , such that exp is one-to-one
on E j =π−1(K )∩Ni ∪Z . For if not, there exists for all j , vectors v j and w j such thatexp(v j ) = exp(w j ). By the above, v j and w j are non zero, so they must lie inπ−1(K )∩Ni ⊂π−1(K )∩N . Taking subsequence and using the continuity of the exponential,we reach a contradiction. By second countability, P contains an increasing sequenceof compact sets Ki such that Ki ⊂ intKi+1 and
⋃Ki = P . For K1, we construct Ei as
above. The rest of the proof follows by an induction argument left as an exercise.
8 Local geometry
8.1 Two parameters maps
We start by some preliminaries on two parameter maps.
Definition 8.1. Let D ⊂ R2 be open and such that D intersected with any vertical orhorizontal line is an interval (possibly empty)26. A two parameter map is a smoothmap
κ : D → M
(u, v) → κ(u, v).
κ is thus composed of two intertwined family of curves
• For any fixed v0 such that v = v0∩D 6= ;, the u-parameter curve v = v0 of κis u → κ(u, v0).
• For any fixed u0 such that u = u0∩D 6= ;, the v-parameter curve u = u0 of κis v → κ(u0, v).
26A wedge (kind of boomerang) obtained from instance from the points (10,0), (0,0), (0,10) and (1,1) isan example of such D which is non-convex.
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At any (u, v) ∈D, we can compute dκ(u,v) : T(u,v)D → Tκ(u,v)M . The maps
(u, v) → κu := dκ(∂u) ∈ Tκ(u,v)M
and(u, v) → κv := dκ(∂v ) ∈ Tκ(u,v)M
are then vector fields on the map κ. Moreover, one has of course that κu(u0, v0) co-incides with the tangent vector of the v = v0 curve at u0 and κv (u0, v0) is the tangentvector of the u = u0 curve at v0.
If the image of κ lies in the domain of a coordinate system (xα) then its coordi-nate functions κα = xα α are smooth real functions on D. Moreover,
κu = ∂u(κα)∂xα , κv = ∂v (κα)∂xα .
Let Z be a smooth vector field on the map κ, i.e.
Z : D → T M ,
such that πZ = κ, where π is the canonical projection π : T M → M .We can then define Zu = D Z
∂u the covariant derivative of Z along the u-parametercurve, given in local coordinates by
Zu = ∂u Zα∂xα +ZαΓβ
αδ∂uκ
δ∂xβ ,
and similarly Zv = D Z∂v the covariant derivative of Z along the v-parameter curve.
As in the case of curves, in the particular case that Z = κu , Zu is the accelerationof the u-parameter curve. Moreover,
Lemma 8.1. 1. If κ is a two parameter map,
κuv = κvu .
2. If Z is a vector field on κ, then
Zuv −Zvu = R(κv ,κu)Z
where R is the Riemann curvature tensor.
Proof. Writing κuv in coordinates, it follows that it is symmetric in uv . We leave theproof of the second point as an exercice.
Remark 8.1. The induced covariant derivative on a curve, the induced covariantderivative on a submanifold and the two covariant derivatives for vector fields on atwo-parameter map are in fact all special examples of the induced covariant deriva-tive for sections of the pullback bundle. You can try to write a definition for it as anexercice.
8.2 The Gauss lemma
So far we have seen that given p ∈ M , expp carries rays t → t v into radial geodesicst → γv (t ). Moreover, we computed the differential of the exponential map at 0 ∈Tp M . The following result, called the Gauss lemma is concerned with the differen-tial map of the exponential map at some radial vector. It implies in particular thatorthogonality to radial directions is preserved by the exponential map.
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Lemma 8.2. Let p ∈ M and 0 6= x ∈ Tp M. If vx , wx ∈ Tx Tp M with vx radial, then
Remark 8.2. Here, on the RHS, we evaluate the metric g at p at the vectors vx andwx . Strictly speaking, we should replace them in this expression with their images bythe canonical isomorphism that allows to identify Tw Tp M and Tp M.
Remark 8.3. Recall that given Φ : V →V , for V a vector space endowed with a scalarproduct g , Φ is an isometry if g (dφx (v),dφx )(w) = g (v, w), for all v, w. Thus, theGauss lemma can be viewed as a statement about the exponential map being a par-tial isometry (it preserves certain directions). Note also that in the statement, the dif-ferential of the exponential map is taken at x (hence the notation dx ).
Proof. vx radial means that v = λx (up to the canonical isomorphism) for some λ ∈R and by linearity, we can assume that λ= 1, i.e. v = x and we replace x by v below.
Consider the two parameter map x(t , s) = t (v + sw) in Tp M and its image by theexponential map in M
x(t , s) = expp x(t , s).
We have x(1,0) = v , xt (1,0) = vv and xs (1,0) = wv , hence
xt (1,0) = dv expp (vv ), xs (1,0) = dv expp (wv ).
The statement of the lemma is then
g (xt (1,0), xs (1,0)) = g (v, w).
By definition, the curve t → x(t , s) is a geodesic with initial velocity v + sw . Hencext t = 0 and g (xt , xt ) = g (v + sw, v + sw), for all t .
By Lemma 8.1, xt s = xst . Thus,
∂t g (xt , xs ) = g (xt , xst ) = g (xt , xt s ) = 1
2∂s g (xt , xt ).
Since g (xt , xt ) = g (v, v)+ s2g (w, w)+2sg (w, v), we have
1
2∂s g (xt , xt )(t ,0) = g (w, v).
On the other hand, x(0, s) = expp (0) = p, for all s, so xs (0,0) = 0 and g (xt , xs )(0,0) =0. Thus, we have
g (xt , xs )(t ,0) = t g (v, w),
which gives the lemma when evalutated at t = 1.
Exercise 8.1. Assume here that dim M = n ≥ 2 and that (M , g ) is a Riemannian man-ifold. In Lemma 3.13, we introduced normal coordinates, by fixing an orthonormalbasis of Tp M and using the exponential map. Equivalently, we introduced a spe-cial coordinate system on Tp M, the one given by the orthonormal basis, and thenused that expp is a diffeomorphism to move these coordinates to M. Instead, fixagain an orthonormal basis, and consider polar coordinate on Tp M adapted to thisbasis, r,ω1, ..,ωn−1, where ω := (ω1, ..,ωn−1) are standard coordinates on Sn−1 andr (v) = ||v ||(= g (v, v)1/2). Denote such a coordinate system by ξ. For instance, if n =3, if (x, y, z) is the coordinate system associated to the orthonormal basis, for any
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v ∈ Tp M, ξ(v) = (r,θ,φ), where (θ,φ) denotes the usual coordinates on S2, so that(z = r cosθ, y = r sinθ sinφ, x = r sinθcosφ
). Again, these coordinates can then be trans-
ferred to M via the exponential map. We denote them in this exercise by the sameletters.
1. Check that r → expp (ξ−1(r,ω)) is a radial geodesic (this follows just by defini-tion).
2. Check that the Gauss lemma translates into
g (∂r ,∂r ) = 1, g (∂r ,∂ωi ) = 0.
In other words, the metric takes the form
g = dr ⊗dr + gi j dωi ⊗dω j ,
the points being that there are no cross terms, and that gr r = 1.
3. What fails in the Lorentzian case ?
8.3 Hyperquadrics
Denote by q the line element at p, i.e. the map
q : v ∈ Tp M → gp (v, v).
Let q = q exp−1p . (In the following, we will often use a upper ˜ to denote objects
defined on Tp M . )
Let U be a normal neighborhood at p and U = exp−1p (U ).
At p, we can consider the level sets of q . These are called hyperquadrics. Sincewe can choose coordinates such that, at p, g = η for g Lorentzian or g = δ for g Rie-mannian, we can identifies the level sets of q with the level sets of the line elementof the Minkowski space or the Euclidean space. Note that since the metric is non-degenerate, d qv 6= 0 at any v 6= 0, so that, for any c ∈ R such that q−1(c) 6= ;, thehyperquadrics q−1(c) are honest hypersurfaces of Tp M if c 6= 0. If g is Lorentzian,then q−1(0) \ 0 is a (non-connected) hypersurface of Tp M \ 0 (we exclude v = 0since it belongs to the level set q = 0 but d q0 = 0), called the null cone at p.
Similarly, we can consider, for c 6= 0,
q−1(c) = r ∈U : q(r ) = c,
called local hyperquadrics. Now, by definition,
q(r ) = q exp−1p (r ),
for all r ∈ U , so if v ∈ q−1(c)∩ U , q(expp (v)) = c and q−1(c) is the image by the
exponential map of q−1(c) ∩ U . In the Lorentzian case, the local null cone at p,denotedΛ(p), is by definition27 Λ(p) = q−1(0) \ p.
Let c 6= 0 such that q−1(c) 6= ;. Since expp : q−1(c)∩U → q−1(c)∩U by construc-
tion, for any v in q−1(c)∩ U , we have
dv expp : Tv
(q−1(c)∩ U
)→ Texpp (v)
(q−1(c)∩U
).
27The point p has to be removed so that these form regular submanifolds of M .
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Excluding 0 ∈ Tp M and p ∈U , this is also true for c = 0, i.e. for the null cone andthe local null cone.
Recall the definition of the position (also called scaling) vector field P . It is avector field of Tp M , i.e P ∈ Γ(Tp M) and it is given in coordinates by
P (v) = vα∂vα ,
for v ∈ Tp M .On the other hand, since q is a function on Tp M , its differential d q ∈Λ(Tp M) is
a one-form of Tp M and its gradient a vector field on Tp M .For v ∈ Tp M , q(v, v) = gαβvα so that
d qv = 2gαβvαd vβ,
and multiplying by g−1, we obtain the following lemma.
Lemma 8.3. We havegrad q = 2P .
As a consequence, the position vector field is orthogonal to every hyperquadric and,in the Lorentzian case, is both tangent and orthogonal to the null cone q−1(0) \ 0.Finally, by construction, the position vector field is radial (trivial).
Using the exponential map, we can then define the local position vector field atp as follows. Let q ∈U and v = exp−1
p (q). Then define P as
P (q) = dv expp (P ).
From the Gauss lemma, it follows that
Corollary 8.1. The local position vector field P at p is orthogonal to every local hy-perquadric of M at p. Furthermore, in the Lorentzian case, P is both orthogonal andtangent to the nullcone q−1(0)\p. Finally, P is radial, that is tangent to every radialgeodesics emanating from p.
Proof. Let c 6= 0 and consider a local hyperquadric q−1(c) 6= ;. Let X be tangent toq−1(c) at some point r . Let v ∈ Tp M be such that expp (v) = r . Now, since we havean isomorphism
dv expp : Tv
(q−1(c)∩ U
)→ Texpp (v)q−1(c),
we can take x ∈ Tv
(q−1(c)∩ U
)be such that dv expp (x) = X . Then,
g (P, X )(r ) = g (dv expp (P ),dv expp (x)) = g (P , x),
by the Gauss Lemma. Since x is tangent to q−1(c) and P is orthogonal to q−1(c) bythe previous lemma, we have g (P, X )(r ) = 0. We leave the rest as an exercice.
Corollary 8.2. Let Dq denotes the gradient vector field of q. Then, Dq = 2P.
Proof. First, we have Dq = 2P from Lemma 8.3. Let r ∈U and v ∈ Tr U . Let w suchthat expp (w) = r and v be such that dw expp v = v . Then,
gr (Dq, v) = v(q) =(dw expp v
)(q) = v
(q expp
)= v(q) = g (Dq , v) = 2g (P , v) = 2g (P, v).
where in the last step we used the Gauss lemma.
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8.4 Convex neighborhoods
Recall the concept of convexity in Euclidean geometry: a set S in an Euclideanspace is convex if and only for all 0 ≤ s ≤ 1 and all points p, q ∈S , p(1− s)+ sq ∈S .Let us try to translate this in an arbitrary semi-Riemannian manifold. Given a setS and p, q ∈ S , the replacement for the curve s → ps + (1− s)q should clearly be ageodesic s → γ(s) from p to q . This motivates the following definition.
Definition 8.2. An open set C is convex provided it is a normal neighborhood of eachof its points.
In particular, for any two points p, q of C , there is a unique geodesic segmentσpq : [0,1] → M from p to q that lies entirely in C (but contrary to the Euclidean orMinkowskian case, there could well be others that leave C ).
The aim of this section will be to prove
Proposition 8.1. Let M be a semi-Riemannian manifold. Then for any p ∈ Tp M,there exists a convex set C containing p.
Before we prove this proposition, let us introduce some extra material concern-ing the exponential map, viewed this time as a mapping defined on T M instead ofthe individual tangent plane. More precisely, let π : T M → M be the canonical pro-jection. Let D be the set of all vectors v ∈ T M , such that the curve γv defined asbefore as the unique maximal geodesic starting at π(v) with initial tangent vector vis defined on [0,1] (i.e. its maximal time of existence is larger than 1). Define as wellDp =D∩Tp M . (Here we identify Tp M with a subset of T M , which if T M is definedas the disjoint union of the Tp M , is the subset p×Tp M).
Lemma 8.4. D is open in T M and Dp is an open set of Tp M that is starshaped around0.
Proof. That D is open follows from standard ode theory (in particular continuousdependance with respect to the initial data). That Dp is starshaped around 0 is aconsequence of the fact γv (st ) = γsv (t ), for 0 ≤ s, t ,≤ 1.
Let us now define E as
E : D ⊂ T M → M ×M
v = (p, vp ) →(p(=π(v)),expp (vp )
)Recall that a smooth map between two manifolds is called non-singular at some
point if its differential at that point is injective28 (one-to-one). We have
Lemma 8.5. Let p ∈ M and w ∈ Dp . If expp : Dp → M is non-singular at w, thenE : D → M ×M is non-singular at w.
Proof. Suppose dw E(v) = 0 for v ∈ Tw (T M). We must show that v = 0. Let π :T M → M denote the canonical projection and let π1 be the projection of M × Mon its first factor. Then π1 E = π, so that dπw (v) = dπ1(dw E(v)) = 0. It followsthat v is vertical, that is tangent to Tp M . Since Dp is open, we can find a curve inσ : I → Dp , such that σ′(0) = v , σ(0) = w . The result then follows since, for all t ∈ I ,E(σ(t )) = (p,expp (σ(t ))) and we have assumed that expp is non-singular at w .
28A regular point on the other hand is one where the differential is onto.
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For any p ∈ M , it then follows from the inverse function theorem and Proposition3.8 that E maps some open neighborhood of (p,0p ) in T M diffeomorphically ontoan open neighborhood of (p, p) ∈ M×M . We now proceed to the proof of Proposition8.1.
Proof. Let ξ= (xα) be a normal coordinate system on a neighborhood of p ∈ M . LetN (δ) be the set of points q such that∑
α|xα(q)|2 < δ.
For δ small enough, this set defines a normal neighborhood of p which is diffeo-morphic to an open ball. (Note that Nδ is clearly open and contains p. Thus, it is aneighborhood of p and clearly, exp−1
p (Nδ) is a neighborhood of 0 ∈ Tp M . Using thedefinition of normal coordinates, it is not hard to check that this neighborhood isstarsphaped around 0. )
By the previous lemma, if δ is small enough, then E is a diffeomorphism from aneighborhood N containing (p,0) ∈ Tp M onto N (δ)×N (δ). Moreover, for δ smallenough, the tensor field whose components are given by δαβ − Γγαβδγρxρ can be
assumed to be positive definite. We claim that it then follows that N (δ) is a normalneighborhood of each of its points q ∈ N (δ).
Let Nq =N ∩Tq M . By construction, E|Nq is a diffeomorphism onto q×N (δ),hence [expq ]|Nq is a diffeormorphism onto N (δ). It remains to show that Nq is star-shaped about 0.
Let r ∈ N (δ), r 6= q and let v = E−1(q,r ). Then, v ∈ Nq and σ = (γv )|[0,1] is ageodesic from q to r . If σ lies in N (δ) then t v ∈Nq for all 0 ≤ t ≤ 1 and hence Nq isstarshaped around 0.
For any k ∈ N (δ), let n(k) = ∑α |xα(k)|2. We consider the function n σ and dif-
ferentiate twice, writing xα for xα(σ) and using the geodesic equation:
d 2n σd t 2 = 2
(δβαxβ
d 2xα
d t 2 +[
d xα
d t
]2)
= 2(δαβ−δγρxρΓγ
βα
) d xα
d t
d xβ
d t> 0,
where we have used the geodesic equations to reach the last line. Thus, the functionn σ is (strictly) convex and in particular, for any t ∈ [0,1], n σ(t ) ≤ (1− t )n σ(0)+t .n σ(1) < δ.
Remark 8.4. In the above proof, we introduce a Riemannian metric (the Euclideanmetric associated to the normal coordinates) on our neighborhood. This is a rare in-stance where one can draw conclusions in arbitrary semi-Riemannian manifolds byforcing a Riemannian structure.
If p and q are points of a convex set C and σpq is the geodesic in C from p to q(unique up to parametrization), the displacement vector −→pq is by definition
−→pq :=σ′pq (0) ∈ Tp M .
Lemma 8.6. Let C be a convex (open) set. Then, the map ∆ : C ×C → T M sending(p, q) to −→pq is smooth.
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Proof. Recall the map E : D → M ×M . We have that D is open and that E is smooth.In particular, since C ×C is open in M ×M , it follows that E−1(C ×C ) is open. Oneeasily check that E−1(C ×C ) =∆(C ×C ) and then that ∆ is the inverse map of
E :∆(C ×C ) →C ×C .
Exercise 8.2. 1. ConsiderS1 ⊂R2. Find two convex sets whose intersection is nonconnected, hence in particular non convex.
2. Prove that if U ,V are two convex sets included into a convex set W , then U ∩Vis a convex set if non-empty.
Definition 8.3. A convex covering is a covering of M by convex open sets such that isU and V are elements of the covering, then their intersection U ∩V is convex if notempty.
Let us end this section with the following lemma, whose proof is left as an exer-cice.
Lemma 8.7. Given any open covering C of M, there exists a convex covering R suchthat each element of R is contained in some element of C . In other words, the convexcovering is a refinement of the original cover.
Proof. (sketch) Let C be an open covering of M . Recall that M is metrizable, so let dbe a metric on M which is compatible with the topology of M . For any x ∈ M , let Nx
be a convex neighborhood containing x. Shrinking Nx is necessary, we may assumethat Nx is bounded (for d), has compact closure and is contained in some element ofC . Then, let Vx be a convex neighborhood of x, such that diam(Vx ) < 1/3d(x,∂Nx ).The set of all such Vx , x ∈ M , is then a covering of M , such that each Vx is containedin some element of C and is convex. Moreover, if Vx ∩Vy 6= ;, assume wlog thatdiam(Vx ) ≥ diam(Vy ) then by construction,
d(x, y) ≤ 2diam(Vx ) ≤ 2/3d(x,∂Nx ).
Thus, y ∈ Nx and then Vy ∈ Nx . Thus, Vx ∩Vy ∈ Nx so that Vx ∩Vy is convex fromExercise 8.2.2.
8.5 Arc lengths
Definition 8.4. Let α be a piecewise smooth curve defined on a closed interval [a,b].We define its arc length as
L(α) =∫ b
a|g (α′,α′)|1/2d s.
A reparametrization function h : [c,d ] → [a,b] is a piecewise smooth functionsuch that either h(c) = a, h(d) = b (h is orientation preserving), or h(c) = b, h(d) = a(h is orientation-reversing). If its derivative does not change sign, h is monotone.
Lemma 8.8. The length of a piecewise smooth curve segment is invariant under achange of monotone reparametrization. Moreover, if |g (α′,α′)| > 0, there is a strictlyincreasing reparametrization function h such that β=αh has |g (β′,β′)| = 1.
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A curve such that |g (α′,α′)| = 1 is said to have unit speed or arc length parametriza-tion.
Let U be a normal neighborhood of p ∈ M . The function
r := k ∈U → r (k) =∣∣∣g (
exp−1p (k),exp−1
p (k))∣∣∣1/2
is the radius function at p.
Exercise 8.3. Let (xα) be normal coordinates at p. Give an expression for r (q) in termsof the coordinates of q.
Lemma 8.9. Let r be the radius function on a normal neighborhood U of p ∈ M. If σis the radial geodesic from p to q ∈U , then L(σ) = r (q).
Proof. If v =σ′(0), then v = exp−1p (q). Since g (σ′,σ′) is constant along σ,
L(σ) =∫ 1
0|g (σ′,σ′)|1/2 = |g (v, v)|1/2 = r (q).
8.6 The Riemannian case
We assume here that g is a Riemannian metric. Then, v → r (v) = g (v, v)1/2 is a normon each tangent plane and is smooth away from v = 0. so that, given a normal neigh-borhood U of p, the radius function at p,
r = r exp−1p
is smooth expect at p. Let q(v) = g (v, v) and q = q exp−1p and denote by P the local
position vector field, i.e. P = d expp P , with P given in local coordinates by
P = vα∂vα .
Then, for all q = expp (v),
|g (P,P )|1/2(q) = g (d expp P ,d expp P )1/2(q) = g (P , P )1/2 = g (v, v)1/2 = r (q).
Hence, U = P/r is the outward unit radial vector field (outward because r increasealong the integral curves of U , radial because P is radial) and U is normal to all thehyperspheres at p, i.e. the hypersurfaces of constant r . Since r = q1/2, it follows fromCorollary 8.2 that
g r ad r = P
r=U
on U \ p.
Proposition 8.2. Let (M , g ) be a Riemannian manifold and U be a normal neigh-borhood of p. If q ∈ U , then the geodesic σ : [0,1] → U with σ(0) = p, σ(1) = q is theunique (up to monotone reparametrization) shortest (piecewise smooth) curve in U
from p to q.
Remark 8.5. Note that here everything is constrained to stay in the neighborhood U .
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Proof. We must prove that if α : [0,b] → U is another curve joining p to q , thenL(α) ≥ L(σ) with equality if and only ifα=σh for some monotone h. We will proveit in the case of smooth curves for simplicity, but modulo easy modifications theproof works for piecewise smooth curves.
First, restricting the vector field U = P/r to α, we have
α′ = g (α′,U )U +N ,
where N is a vector field on α orthogonal to U for all t > 0. At t = 0, the vector fieldU is a priori not well defined, but since g (U ,U ) = 1, its components are uniformlybounded in any local system of coordinates. (In fact, U can be extended continu-
ously to α′(0)r (α′(0)) .)
Then,
g (α′,α′)1/2 = [g (α′,U )2 + g (N , N )]1/2 ≥ |g (α′,U )| ≥ g (α′,U ).
On the other hand, from grad r =U , we have g (α′,U ) = drαd t . Hence,
L(α) =∫ b
0g (α′,α′)1/2d s ≥
∫ b
0g (α′,U )d s = r (q) = L(σ).
Moreover, in the equality case, then we have N = 0, for all t > 0 as well as |g (α′,U )| =g (α′,U ), which implies that
dr αd t
≥ 0.
Thus, α′ = drαd t U , which implies that α is a monotone reparametrization of an in-
tegral curve of U . Since U is radial, the integral curves of U are geodesics passingthrough p, so it follows that α is a monotone reparametrization of a geodesic fromp to q , which by uniqueness has to be σ. In fact, α(t ) =σ(r α(t )/r (q)).
8.6.1 The Riemannian distance
Definition 8.5. For any points p and q of a connected Riemannian manifold M, theRiemannian distance from p to q is the infimum of all the lengths of all piecewisesmooth curves joining p to q.
Remark 8.6. Recall that any connected manifold is path connected, so that the abovedefinition makes sense.
If p ∈ M and ε> 0, we define Bε(p) to be the open ball of size ε for the Riemanniandistance d , i.e.
Bε(p) = q ∈ M : d(p, q) < ε.
Note that for the moment, we do not know that this is actually open for the topol-ogy of M .
Proposition 8.3. Let p ∈ M.
1. For any ε> 0 sufficiently small, Bε(p) is normal (in particular open).
2. For q in a normal neighborhood of the form Bε(p), the radial geodesic σ fromp to q is the unique shortest curve in M from p to q. In particular,
L(σ) = r (q) = d(p, q).
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Remark 8.7. The point of the second item above is that we are considering all possiblegeodesics, not only those constrained in the normal neighborhood.
Proof. Let V be a normal neighborhood of p ∈ M and U = exp−1p (V ). For ε> 0 suffi-
ciently small, U contains the starshaped open set
Bε(0) = v ∈ Tp M : g (v, v)1/2 < ε.
Thus, N := expp (Bε(0)) is also a normal neighborhood of p and by construction N ⊂Bε(p) (cf Lemma 8.9). Moreover, if q ∈ N , then by the previous lemma, the radialgeodesicσ from p to q is the unique shortest curve in N from p to q and L(σ) = r (q).Since L(σ) < ε, it suffices to prove that if α is a curve in M starting at p and leavingN , then L(α) ≥ ε. In particular, this shows that if q ∉ N , then d(p, q) ≥ ε and hencethat N = Bε(p). To prove the claim, note that since α leaves N , it meets every sphereS(a) of constant r equal to a, for all a < ε. If αa is the shortest initial segment of αfrom p to S(a), thenαa lies in N . Hence L(αa) ≥ a by the previous lemma. Since thisholds for all a < ε, the claim is proved.
Proposition 8.4. For a connected Riemannian manifold M, the Riemannian dis-tance function d : M ×M →R defines a metric on M, i.e. it safisfies
1. for all p, q ∈ M, d(p, q) = 0 if and only if p = q.
2. for all p, q ∈ M, d(p, q) = d(q, p),
3. for all p, q,r ∈ M, d(p, q) ≤ d(p,r )+d(r, q).
Moreover, d is compatible with the topology of M.
Proof. The symmetry property is trivial. Assume that d(p, q) = 0, since M is Haus-dorff, there is a normal neighborhood of p that does not countain q and from the(proof of the) previous proposition, there exists a open ball around p of radius ε> 0that does not contain q . Thus, d(p, q) > ε> 0.
For the triangle inequality, let ε > 0 and choose α a curve from p to q and β acurve from q to r , such that L(α) < d(p, q)+ε and L(β) < d(q,r )+ε. Joining α and βgives a curve of length at most L(α)+L(β)+2ε. In other words, the triangle inequalityfollows from the definition of infimum of a set and the fact that given curves fromp to q and q to r , we can get a curve from p to r whose length is the sum of theprevious length.
Finally, let p ∈ M and consider q ∈ Br (p), r > 0. We have d(p, q) < r and so,by the triangle inequality, for any r ′ such that 0 < r ′ < r −d(p, q), the ball Br ′ (q) iscontained in Br (p). Since for r ′ small enough, we know from the previous propo-sition that Br ′ (q) is open, it follows that Br (p) is open for the topology defined byM . Conversely, we have already proven that any open set for the topology of M con-tains a small enough ball around any point of that set. Thus, the two topology arecompatible.
A minimizing segment from p to q is by definition a curve from p to q that re-alizes the Riemannian distance function. Note that there could be several or nonesuch segments. One has
Corollary 8.3. A minimizing segmentσ from p to q is a monotone reparametrizationof a smooth geodesic segment from p to q.
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Proof. The domain I of α can be decomposed into subintervals Ii such that eachαi = α|Ii lies a convex open set. Since α is minimizing, each αi is also minimizing(the restriction of a minimizing segment is a minimizing segment). Since each αi
lies in a convex set, it follows that it is a monotone reparametrization of a geodesicwhich we can assume to be of unit speed σi . Joining this gives a possibly brokengeodesicσ from p to q and joining the reparametrization gives thatα is a monotonereparametrization of σ. Moreover, we claim that in fact σ is unbroken, i.e. it is asmooth geodesic. For, consider two geodesic segmentsσi ending at r and a geodesicsegment σi+1 starting at r . Since α is minimizing, then σ is minimizing. Thus, therestriction of σ to any sufficiently small neighborhood of r is minimizing. But thisimplies that this restriction is a monotone reparametrization of a smooth geodesic.Since we had assume σ to be parametrized by arc length, it follows that in fact σ is asmooth geodesic, which concludes the proof.
8.6.2 The Hopf-Rinow Theorem
The aim of this section is to prove the following theorem.
Theorem 8.1 (Hopf-Rinow (1931)). For a connected Riemannian manifold M thefollowing are equivalent
(MC) Metric completeness: (M ,d) is a complete metric space, i.e. every Cauchy se-quence is convergent.
(GC1) Geodesic completeness from one point: There exists a point p ∈ M such thatevery geodesic starting at p is defined on R.
(GC) Geodesic completeness: M is geodesically complete.
(HB) Heine-Borel property: Every closed bounded subset of M is compact.
With the notations of the theorem, we start by proving
Lemma 8.10. If (GC1) holds, then for any q ∈ M, there exists a minimizing geodesicsegment from p to q.
Proof. Consider a normal ball Bε(p), where ε > 0 small enough. If q ∈ Bε(p), thenthe claim follows from Proposition 8.3, thus we assume that q ∉ Bε(p). Let r bethe radius function at p. For 0 < δ < ε, the level set r = δ is an (n −1) dimensionalsphere in Bε(p) denoted Sδ. In particular, Sδ is compact. The function s → d(s, q) iscontinuous on Sδ, hence reach its minimum at some point m ∈ Sδ. We claim that
d(p,m)+d(m, q) = d(p, q).
To prove this, let α : [0,b] → M be any curve from p to q . The function τ(s) : s →d(α(s), p) is continous and satisfies τ(0) = 0, τ(b) = d(p, q) > ε. By the intermediatevalue theorem, there exists an a, such that d(α(a), p) = δ. By definition, α(a) ∈ Bε(p)and in fact α(a) ∈ Sδ, cf Proposition 8.3. Moreover, 0 < a < b. Let α1 and α2 be therestriction of α to [0, a] and [a,b] respectively. Then,
L(α1) ≥ δ= d(p,m)
by the minimizing property of radial geodesics and
L(α2) ≥ d(m, q)
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by the definition of m.Thus,
L(α) = L(α1)+L(α2) ≥ d(p,m)+d(m, q).
Hence,d(p, q) ≥ d(p,m)+d(m, q).
The reverse inequality is just the triangle inequality, so the claim is proven.Now, let γ : [0,+∞) → M be the unit speed geodesic whose initial segment runs
radially from p to m. (Note that γ is defined on [0,+∞) since we assumed (GC1).)Let d = d(p, q) and let T be the set of t ∈ [0,d ] such that
t +d(γ(t ), q) = d .
It suffices to show that d ∈ T . For, it then follows that d(γ(d), q) = 0, which impliesthat γ(d) = q . Since γ has unit speed, L(γ[0,d ]) = d , so that γ[0,d ] is indeed a minimiz-ing geodesic from p to q .
Note first that γ[0,t ] is minimizing for any t ∈ T , since its length is t and
d ≤ d(p,γ(t ))+d(γ(t ), q) = d(p,γ(t ))+d − t ,
which implies that d(p,γ(t )) ≤ t and thus that d(p,γ(t )) = t .By continuity, T is closed. Moreover T is non-empty by the first part of the proof.
Thus, it contains a largest number t0 ≤ d . Assuming t0 < d , we deduce a contractionas follows. In a normal neighborhood U ′ of γ(t0), repeating the first part of the proofproduces a unit speed radial geodesicσ : [0,δ′] →U ′ from γ(t0) to m′ ∈U ′ such that
δ′+d(m′, q) = d(γ(t0), q). (26)
Moreover, because t0 ∈ T , we have
t0 +d(γ(t0), q) = d ,
so that from (26),t0 +δ′+d(m′, q) = d .
Since d ≤ d(p,m′)+d(m′, q) and since t0 +δ′ is the length of a curve from p to m′,we must have t0+δ′ = d(p,m′). Thus, the sum of γ[0,t0] andσ is minimizing, so fromCorollary 8.3 and the fact that they have unit speed (so the parametrization is fixed),the sum is a smooth minimizing geodesic. This means that m′ = σ(δ′) = γ(t0 +δ′),so that
t0 +δ′+d(γ(t0 +δ′), q) = d(p,m′)+d(m′, q) = d ,
hence t0 +δ′ ∈ T a contradiction.
For the proof of the Hopf-Rinow theorem, we need the notion of an extendiblecurve.
Definition 8.6. A piecewise smooth curve α : [0,B) → M is extendible provided
lims→B ,s<B
α(s)
exists. We then say that α has a continuous extension α : [0,B ] → M and q = α(B) iscalled an endpoint of α.
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Equivalently, there exists a q ∈ M such that for every sequence si → B , α(si ) → q .An extendible curve need not have a piecewise smooth extension however, it does inthe special case of integral curves, as a consequence of the Cauchy-Lipshitz theoremand the standard continuation criterion for odes. Note that this above definitionmakes sense for manifolds in general (it is not restricted to Riemannian manifolds).
Exercise 8.4. 1. Let (M , g ) be semi-Riemannian manifold. With the above nota-tion, prove that if α is a geodesic, then it is extendible if and only if it extendibleas a smooth geodesic. (In the Riemannian case, this follows from Cauchy-Lipschitzand the standard continuation criterion for odes, for a solution in the generalcase, see for instance Lemma 8, p.130 in [O’N83].)
2. Let X be vector field on M that never vanishes (equivalently its flow has no fixedpoint). Prove that the maximal integral curves of M are inextendible. In partic-ular, the maximal integral curves of a globally causal vector field are all inex-tendible.
Proof of the Hopf-Rinow Theorem 8.1.
(MC) =⇒ (GC). Let γ : [0,b) be a unit speed geodesic. Assume that b <+∞. If (ti ) is a sequencein [0,b) converging to b then (γ(ti )) is Cauchy, since
d(γ(ti ),γ(t j )) ≤ |ti − t j |.Hence,(γ(ti )) converges to some q ∈ M . Moreover, if (si ) is another such se-quence, then we still have d(γ(ti ),γ(s j )) ≤ |ti − s j |, so that the limit points isunique. It follows that γ is continuously extendible and it follows from theabove exercice that in fact it is extendible as a smooth geodesic. Hence, everymaximal geodesic is defined on (−∞,+∞) i.e. M is geodesically complete.
(GC) =⇒ (GC1) Trivial.
(GC1) =⇒ (HB) Let p be as in (GC1). Let A be closed bounded subset of M . Note that bydefinition, A bounded means that d is bounded on A × A and by the triangleinequality, it follows that q ∈ A → d(q, p) is bounded on A.
Let q ∈ A. By Lemma 8.10, there exists a minimizing geodesic σq defined on[0,1] from p to q . Then g (σ′
q (0),σ′q (0))1/2 = L(σq ) = d(p, q) < r , for some r > 0
independent of q . In other words, σ′(0) lies in the compact ball Br of Tp M .Since expp (Br ) is compact and contains the closed set A, A is compact.
(HB) =⇒ (MC) Always true in metric spaces, so left as an exercice.
As a consequence of the Hopf-Rinow theorem (and the lemma), we have
Corollary 8.4. If a connected Riemannian manifold is complete then any two of itspoints can be joined by a minimizing geodesic segment.
Corollary 8.5. Any compact Riemannian manifold is complete.
Proof. Property (HB) holds trivially in this case.
Finally, let us recall Whitney’s embedding theorem29 and obtain yet another corol-lary of the Hopf-Rinow theorem (combined with Whitney’s embedding theorem).
29Actually, Whintney proved that M can be emdedded into R2n , but the 2n+1 version is easier to proveand we do not care here about the dimension of the target space.
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Theorem 8.2. Let M be a smooth connected manifold of dimension n. Then, thereexists an embedding of M into R2n+1 such that the image under the embedding is aclosed subset.
For a proof, see for instance Theorem 10.8, p92 in [Bre97].
Corollary 8.6. Let M be a smooth connected manifold. Then, there exists a completeRiemannian metric on M.
Proof. Let φ : M → R2n+1 denotes an embedding as in Whitney’s theorem and giveM the pullback metric by the embedding. Let dM be the corresponding Riemanniandistance. Then, closed and bounded set of M with respect to d are easily seen tobe compact (use for instance that dR3 (φ(p),φ(q)) ≤ dM (p, q), p, q ∈ M) ), so thatproperty (HB) is satisfied.
For an alternative, more constructive, proof, see Lemma 11.1, p111 in [Rin09].
8.7 The Lorentzian case
Many (most) of the previous properties proved in the case of Riemannian manifoldsdo not hold for Lorentzian manifolds. As a start, g does not defines a norm on eachtangent space, so in particular, for any ε > 0, the set of all v , such that |g (v, v)| ≤ ε
is non compact. The bottom line is that, yes there are analogies between the Rie-mannian and Lorentzian cases (for instance, we will soon see a local maximizingproperty of timelike geodesics analogous, including the proof, to the minimizingproperty of geodesics in Riemannian geometry), but overall Lorentzian manifoldshave their own, more complicated, geometry.
We now assume for the rest of this section that (M , g ) is a smooth Lorentzianmanifold.
8.7.1 Timecone and time-orientability
Recall that v ∈ Tp M 6= 0 is called spacelike, null or timelike depending on whethergp (v, v) > 0, = 0 or < 0 and that the set causal vectors is the union of the null andtimelike vectors. Similarly, a submanifod is also called spacelike, null or timelike de-pending on whether corresponding the induced metric is Riemannian, degenerateor Lorentzian. In particular, this applies to smooth curves. Moreover, if z ∈ Tp M istimelike, then z⊥ is spacelike (in particular, by convention, 0 ∈ Tp M is spacelike).
The proof of the following lemma is left as an exercise.
Lemma 8.11. [Reverse Cauchy-Schwarz inequality] Let v, w be timelike vectors. Then,
(g (v, w))2 ≥ |g (v, v)| . |g (w, w)|.
Let Tp denotes the set of all timelike vectors at some p ∈ M . On Tp , we considera binary relation defined as follows. We say for v, w,∈ Tp that v C w if and only ifg (v, w) < 0.
Lemma 8.12. 1. The relation C is an equivalence relation and there are exactlytwo equivalent classes, referred to as timecones.
2. The two timecones are open and connected, so that Tp has two connected com-ponents.
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3. The two timecones are convex sets.
4. The map v ∈Tp →−v maps one timecone into the other.
Proof. Clearly, C is a symmetric binary relation since g is symmetric. That C isreflexive follows from the definition of a timelike vector. Moreover, if g (v, w) < 0and g (w, x) < 0, then since w is timelike, it follows that w 6= 0, and we can findan orthonormal basis (ei ) such that w = |g (w, w)|1/2e0 where g (e0,e0) = −1. Then,g (v, w) < 0 implies that v = v0e0+v i ei , with v0 > 0 and v0 = |v0| > (v i v jδi j )1/2 sincev is timelike and similarly for x. Then,
g (x, v) =−v0x0 + v i x jδi j
so that the result follows by the Cauchy-Schwarz inequality. We leave the rest as anexercise.
With the notion of timecones, we can extend the notation of timelike, null orcausal to curves which are only piecewise smooth. For a piecewise smooth curve,we say that it is timelike if its tangent vector at every non-break point is timelike andif all its half tangents (left and right ones) are all timelike and points in the sametimecone. Similarly for a null or causal curve. Note that by definition, a causal curveis a "regular" curve, in the sense that its tangent vector at every non-break point isnonzero, nor does any all its half-tangents.
A priori, there is no intrinsic way to distinguish one timecone from the other. Toremediate this, we need an extra structure.
Definition 8.7. (M , g ) is said to be time-orientable if there exists a global timelikevector field T on M. A choice of time-orientation is the prescription of such a globaltimelike vector field T . We then say that (M , g ) is time-oriented by T .
Remark 8.8. The standard notion of orientation for manifolds and the notion oftime-orientation for a Lorentzian manifold are unrelated, cf [O’N83], p145.
Remark 8.9. As for the standard orientation, every Lorentzian manifold admits adouble cover30 that is time-orientable, cf [O’N83, Chapter 7].
In the remainder all the Lorentzian manifolds that we will consider will be time-orientable. In fact, we will give a name to those.
Definition 8.8. A spacetime is a triple (M , g ,T ) where (M , g ) is a connected Lorentzianmanifold (of dimension n ≥ 2) and T is a global timelike vector field on M.
For simplicity31, on top of being time-orientable, we will assume also that ourspacetimes are orientable in the rest of the lectures. (Thus, for us, all spacetimeswill be orientable by assumption, and we may omit to say it later). Moreover, weoften write (M , g ) instead of (M , g ,T ).
A causal vector v ∈ Tp M is called future-oriented if g (v,Tp ) < 0, i.e. v and Tp
points in the same timecone (if v is null then X does not belong to the timeconeof T ). Similarly for past-oriented and again curves inherit the labels if their tangentvectors at each point have it.
We have easily
30i.e. a smooth map k : M → M , where M is a smooth manifold, such that each p ∈ M has a connectedneighborhood U such that k−1(U ) has two connected components and k maps each of them diffeo-mophically onto U .
31While time-orientability is important to talk about causality, most constructions below and all thephysical interpretations do not need the spacetime manifolds to be orientable.
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Proposition 8.5. Let (M , g ) be a spacetime. Then, there exists a globally defined time-like vector field on M whose flow is defined on M ×R, i.e. the vector field is complete.
Proof. Since M is time-oriented, there exists a globally timelike vector field T1. Us-ing Corollary 8.6, we can introduce a complete Riemannian metric h on M . Let thenT = ||T1||−1
h T1. We claim that T is complete. Indeed, let γ an integral curve of T de-fined on (t−, t+). If t+ <∞ then the integral curve restricted to [a, t+) (a ∈ (t−, t+)) iscontained in a compact set, since it has bounded length for the Riemannian metric.Hence, the curve is extendible through t+ a contradiction.
8.7.2 Local Lorentzian geometry
We have
Lemma 8.13. Let p ∈ M and β : [0,b] → Tp M a piecewise smooth curve at p (withvalues in Tp M not in M!) such that
1. β(0) = 0,
2. α= expp β is well defined on [0,b],
3. α= expp β is timelike.
Then, for all t ∈]0,b], β(t ) is timelike and remains in a single timecone of Tp M.
Remark 8.10. Recall that, for t ∈ [0,b], expp β(t ) is well defined if the geodesic γβ(t )
starting at p and with initial tangent vector β(t ) is defined on [0,1].
Proof. Note first that that if β(t ) is timelike for all t > 0, then β must stay in a singletimecone (since there are only two timecones and there are disconnected).
In the following, we write that property P holds initially for “there exists ε > 0,such that for all t ∈]0,ε], Property P holds”. Assume first that β is smooth. Sinceβ′(0)(=α′(0)) is timelike and β(t ) = tβ′(0)+O(t 2), initially β is in a single timecone,denoted C . Note that the position vector field P on Tp M is by definition radial andpoints in the same direction as β in particular, it is timelike in C . Thus, initiallyg (β′, P ) < 0. Since Dq = 2P , we have
d
d tq β= 2g (β′, P ),
and initially dd t q β is strictly negative.
By the Gauss lemma,g (α′,P ) = g (β′, P )
and so initially, g (α′,P ) < 0.Consider the set E all τ such that for all 0 < t ≤ τ,
d
d tq β(t ) ≤ 0.
By construction, E is a closed in ]0,b] and it is not-empty by the above. Let a = supE .By the above, we know that a > 0. Moreover, on ]0, a], q β is decreasing and thusq β < 0, so β is timelike on ]0, a] and must stay in a single timecone. Thus, Pβ(the position vector field at β) and Pα (by the Gauss lemma) are timelike on ]0, a]and must stay in a single timecone. Since α′ and Pα are timelike, from Lemma 8.11,
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g (α′,P ) cannot vanish and thus, it follows that g (α′,P ) < 0 (since true initially) andthen, as above, that g (β′, P ) < 0 and finally that d
d t q β< 0.Thus E is open and hence E =]0,b]. Hence β is timelike for all t and therefore
stays in a single timecone.Assume now that β is merely piecewise smooth. We know from step 1 that on its
first smooth segment β stays in C . Thus at the first break t1, g (α′(t−1 ),P1) < 0, whereP1 = d expp (Pβ(t1)). Since α is timelike, α′(t+1 ) and α′(t−1 ) must point in the same
direction, and thus g (α′(t+1 ),P1) < 0. It then follows that dd t
(q β)
does not changesign at breaks and the rest of the argument follows similarly.
Exercise 8.5. Prove the analogue of the above lemma replacing timelike and timeconeby causal and causal cone.
The following proposition is the Lorentzian analogue of Proposition 8.2.
Lemma 8.14. Let (M , g ) be a Lorentzian manifold and U be a normal neighborhoodof p. If k ∈ U and there exists a timelike curve from p to k in U , then the radialgeodesic segment σ from p to k is the unique (up to reparametrization) longest time-like curve in U from p to k.
Proof. Let α be a timelike curve in U from p to k. Let r be the radius function of Mat p which we recall is defined as
r (k) =∣∣∣g (exp−1
p (k),exp−1p (k))
∣∣∣−1/2.
In other words r (k) gives the norm of the tangent vector to the geodesic σ goingfrom p to k in an affine time of 1. In yet another words, r (k) gives the norm of theposition vector field. In yet yet another words, r (k) = L(σ), where σ is the uniquegeodesic in U from p to k.
Note that r is smooth apart at p and at the local null cone of p.By the previous lemma β= exp−1
p α lies in a single timecone C . Hence, apart at
t = 0,α lies in a region on which U = Pr is a unit timelike vector field. In particular, U
is timelike at k, soσ is timelike at k and hence just timelike (since g (α′,α′) s constantalong α). Let now
α′ =−g (α′,U )U +N ,
where N is a (spacelike) vector field on α orthogonal to U . Then,
Since Dq = 2P and r = (−q)1/2, we have Dr = −Pr = −U . By the previous lemma,
g (β, P ), hence g (α′,U ) is negative, hence
|g (α′,U )| = −g (α′,U ) = d
d tg (r α).
Consequently,
L(α) =∫
|g (α′,α′)|1/2d t ≤ r (k) = L(σ).
If the length are equal, then N = 0 which implies thatα is a monotone reparametriza-tion of σ.
Again, this lemma holds for piecewise smooth curves (again assuming that, bydefinition, piecewise smooth timelike curves have their half-tangents at break pointspointing in the same timecone).
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9 Deformation of curves
9.1 Variations of a curve
Definition 9.1. A (smooth) variation of a curve segment α : [a,b] → M is a two pa-rameter (smooth) mapping
x : [a,b]× (−δ,δ) → M
(u, v) → x(u, v)
such that x(u,0) =α(u) for u ∈ [a,b].
The u parameter curves of a variation are called longitudinal and the v param-eters curve transerve. The vector field V on α given by V (u) = xv (u,0) is called thevariation vector field of α.
If x(a, s) = α(a) and x(b, s) = α(b) for all s ∈ (−δ,δ), we say that x is a variationwith fixed endpoints (in which case V must vanish at the endpoints).
One easily have
Lemma 9.1. Given any curve α : [a,b] → M and V a vector field on α, there exists avariation of α whose variation vector field is V . Moreover, if V vanishes at the end-points, there exists a fixed endpoints variation of α whose variation vector field is V .
Proof. Just take x(t , s) = expα(t )(sV ). Since α is compact, there exists a δ > 0 suchthat x is well-defined on [a,b]× (−δ,δ).
9.2 From causal to timelike curves
Lemma 9.2. Let α be a causal curve segment in a Lorentzian manifold and let x bea variation of α with variation vector field V . If g (V ′,α′) < 0, then for all sufficientlysmall v > 0, the longitudinal curve αv of x is timelike.
Thus, if v > 0 is sufficiently small, then g (xu , xu) < 0 for all u.
By gluing two smooth curves, one often ends up with a piecewise smooth curve.Thus, it seems natural to consider piecewise smooth variations of piecewise smoothcurves.
Definition 9.2. A variation x of a piecewise smooth curve α : [a,b] → M is a continu-ous map of the form
x : [a,b]× (−δ,δ) → M
(u, v) → x(u, v)
with x(u,0) =α(u) for u ∈ [a,b] and such that for breaks
a = u0 < u1 < .. < uk < b = uk+1
the restriction of x to each set [ui−1,ui ]× (−δ,δ) is smooth.
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A priori, x andα could have different breaks, but we can always add trivial breaksto x and α (i.e. add points to the partition of [a,b] corresponding to α or x whereactually there is no regularity issue).
Note that xv is then smooth on each [ui−1,ui ]× (−δ,δ). In particular, the vari-ation vector field V : u → xv (u,0) is continuous32 on each [ui−1,ui ] and hence on[a,b]. On the other hand, the velocity vector field α′ will generally have a disconti-nuity at each break ui . The discontinuity is measured by the tangent vector
∆α′(ui ) =α′(u+i )−α′(u−
i ) ∈ Tα(ui )M .
Exercise 9.1. 1. Prove that Lemma 9.2 is valid for piecewise smooth curve (withthe usual hypothesisis on the half-tangents at each break points).
2. Prove that the conclusion Lemma 9.2 still holds if g (V ′,α′) ≤ 0 and strictly neg-ative whenever α′ is null.
This implies the following result.
Lemma 9.3. In a Lorentzian manifold, if α is a causal curve that is not a null pre-geodesic (cf Exercise 3.13) from p to q, then there exists a timelike curve from p to qarbitrarily close to α.
Remark 9.1. Here arbitrarily close means "constructed out of a fixed endpoints varia-tion ofα" so that the longitudinal curves are all timelike for v > 0. In particular, givenany open set containing α, one can choose such a timelike curve lying in the open set.
Proof. Without loss of generality, assume thatα is defined on [0,1]. Assume first thatα′(1) is timelike. Let W be the vector field obtained by parallel translation of α′(1)alongα. Then, W andα′ are always in the same causal cone, and since W is timelikeg (W,α′) < 0. Since α′(1) is timelike, by continuity, there exists δ,ε > 0, such thatg (α′,α′) <−δ on [1−ε,1]. Let f be any smooth function on [0,1] that vanishes at theendpoints and such that f ′ > 0 on [0,1− ε]. Set V = f W , then g (V ′,α′) = f ′g (W,α′)is negative on [0,1−ε]. Let x be a fixed endpoint variation with variation vector fieldV . Then for v > 0 small enough, the longitunal curve is timelike on [0,1− ε] by theprevious lemma, and on [1− ε,1] simply by continuity. Now if α is timelike at 0, theabove proof can be repeated and if α is timelike at a nonendpoint s, then we canapply the above result on [0, s] and [s,1]. (Note that if α is regular at s and the aboveconstruction yields a timelike curve that is regular (no break point) at s).
Thus, we are left with the case where α is a piecewise smooth null curve. As-sume first that α is a smooth null curve. Differentiating g (α′,α′) = 0, we obtain thatg (α′′,α′) = 0. Moreover α′′ cannot be colinear everywhere on α to α′, otherwise(cf Exercice) α could be reparametrized to be a null geodesic. Thus the functiong (α′′,α′′) is not identically zero, since orthogonal null vectors are necessarily colin-ear. Let W be a parallel timelike vector field on α in the same causal cone as α′ ateach point, i.e. g (W,α′) < 0. Let V = f W +hα′′, where f and h are functions van-ishing at the endpoints that will be determined below so that g (V ′,α′) < 0. Fromg (α′′,α′) = 0, we have
g (α′′′,α′) =−g (α′′,α′′)32Note that here V is given one value at the break points. This is because since we require x[ui−1 ,ui ] to
be smooth, we have V (ui ) = ∂v x(ui , v), i.e. we can take a partial derivative after evaluation at u1.
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and hence
g (V ′,α′) = f ′g (W,α′)+h′g (α′′,α′)+hg (α′′′,α′)= f ′g (W,α′)−hg (α′′,α")
= g (W,α′)( f ′−hk),
where
k := g (α′′,α′′)g (W,α′)
.
Since k is not identically zero, so there exists a smooth function h vanishing at theendpoints such that ∫ 1
0khdu =−1
Let now f (u) = ∫ u0 (kh +1)du. Then, f vanishes at the endpoints, and f ′ = kh +1 >
kh. Consequently g (V ′,α′) < 0.Now, ifα is a piecewise smooth null curve, any non-pregeodesic segment can be
deformed to a timelike curve, and then by the first argument we can deformed thewhole curve to a timelike curve. Hence, we are left with the case of piecewise nullcurve for which each segment is a pregeodesic. By reparametrization, we might evenassume that each segment is a null geodesic, and without loss of generality, we canconsider only the case of two segments. Let us thus assume that α is a null geodesicon [0, s] and [s,1] with a break at s. Let W be the vector field obtained by paralleltransportation of ∆α′(s) =α′(s+)−α′(s−). We have g (W,α′(s−)) = g (α′(s+),α′(s−)) <0 because at breaks, we have assumed by definition that semi-tangents points inthe same direction, and because if = 0, then there are no breaks (in the sense thatnull orthogonal vectors must be collinear and that a change a parametrization thenallows to remove the break). On the other hand, g (W,α′) is constant on [0, s−] byparallel transport andα being a geodesic on that segment. Thus, g (W,α′) is negativeon [0, s−]. Similarly, it is positive on [s+,1]. Now choose a piecewise smooth functionf that vanishes at the endpoints (i.e. at 0 and 1) and such that f ′ > 0 on [0, s−] andf ′ < 0 on [s+,1]. Then, for V = f W , we have g (V ′,α′) < 0.
Exercise 9.2. Give an example of 2 points in Minkowski space that can be joined by anull curve but cannot be joined by a timelike curve.
9.3 First variation
Let x : [a,b]× (−δ,δ) → M be a variation of a curve segment α. For each v ∈ (−δ,δ),let Lx (v) be the length of the longitudinal curve u → x(u, v). Then Lx (v) is a real-valued function with Lx (0) the length of α. If g (α′,α′) 6= 0 on α, then for δ smallenough, Lx is a smooth function (cf below).
Note that g (α′,α′) > 0 on α means that α is spacelike and g (α′,α′) < 0 meansthat it is timelike. Let ε be the sign of α, i.e. ε= sign(g (α′,α′)) = −1,0,+1.
Lemma 9.4. Let x be a variation of a curve segment α : [a,b] → M with g (α′,α′) 6= 0.If Lx (v) is the length of u → x(u, v) then,
L′x (0) = ε
∫ b
a
g (α′,V ′)|g (α′,α′)|1/2
du,
where ε is the sign of α and V is the variation vector field of x.
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Proof.
Lx (v) =∫ b
a|g (xu(u, v), xu(u, v))|1/2du.
In particular, |g (xu(u, v), xu(u, v))|1/2 is smooth for δ small enough since it does notvanish by continuity. Then,
L′x (0) =
∫ b
a
d
d v|g (xu(u, v), xu(u, v))|1/2du
where |g (xu(u, v), xu(u, v))|1/2 = (εg (xu(u, v), xu(u, v)))1/2 and we compute
d
d v|g (xu(u, v), xu(u, v))|1/2 = 1
2(εg (xu(u, v), xu(u, v)))1/2ε .2g (xu , xuv )
= εg (xu , xvu)∣∣g (xu(u, v), xu(u, v))∣∣1/2
,
using xuv = xvu . Just evaluate at v = 0 to conclude.
Exercise 9.3. Check that the previous lemma holds for x a piecewise smooth variationof α a piecewise smooth curve.
Proposition 9.1 (First variation formula33). Let α : [a,b] → M be a piecewise smoothcurve segment with constant speed c = |g (α′,α′)|1/2 6= 0 and sign ε. If x is a piecewisesmooth variation ofα, with break points at (ui )1≤i≤k such that u0 = a < u1 < .. < ui <.. < uk < uk+1 = b, then
L′x (0) =− ε
c
∫ b
ag (α′′,V )du − ε
c
k∑i=1
g (∆α′(ui ),V (ui ))+[ ε
cg (α′,V )
]b
a.
Proof. We simply split the previous integral at the break points and integrate byparts in u, using that
g (α′,V ′) = d
dug (α′,V )− g (α′′,V ).
Remark 9.2. For a fixed endpoint variation x, i.e. x(a, v) = p and x(b, v) = q for allv ∈ (−δ,δ), the variation vector field V vanishes at a and b and thus the last term alsovanishes.
Corollary 9.1. A piecewise smooth curve α of constant speed c > 0 is an (unbroken)geodesic if and only if the first variation of arc length is zero for every fixed endpointsvariation of α.
Proof. If α is geodesic then α′′ = 0 and there are no breaks. Conversely, supposeL′
x (0) = 0 for every fixed endpoint variation x. First, we show that each segmentα|[ui ,ui+1] is geodesic. It suffices to show that α′′ = 0 for ui < t < ui+1. Let y be anytangent vector to M at α(t ) and let f be a bump function on [a,b] with support
33The first variation formula will help us to find critical points of the length functional, but do notdetermine whether these critical points are actual minimizers. For this, one needs to go beyond the firstvariation, for instance, by computing a second variation formula, at the level of two derivatives of L.
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included in [t−δ, t+δ] ⊂ [ui ,ui+1]. Let Y be the vector field onα obtained by paralleltranslation of y and set V = f Y .
Since V (a) = V (b) = 0, the exponential formula x(u, v) = expα(u)(vV (u)) pro-duces a fixed endpoint variation of α whose variation vector field is V .
Since L′x (0) = 0, and given the support assumption on f , we have
0 =∫ t+δ
t−δg (α′′, f Y )du.
Since this holds for all y,δ> 0 and bump function f , we have g (α′′, y) = 0 for all y , i.e.α′′ = 0. It remains to show that the breaks are trivial, so that α is indeed unbroken.As before, let y be an arbitrary tangent vector at α(ui ) and let f be a bump functionat ui with supp f ⊂ [ui−1,ui+1]. For a fixed variation with vector field f Y , the firstvariation formula gives
0 =−ε/cg (∆α′(ui ), y)
for all y , since α′′ = 0 on each subsegment. Since g is non-degenerate, this impliesthat all the breaks are trivial.
10 Causality and global Lorentzian geometry
Recall that to distinguish between the future and the past, we say that M is time ori-ented if there exists a globally defined timelike vector field, denoted here T . Then,at every p ∈ M , for v ∈ Tp M a causal vector, we say that v is future oriented ifgp (v,Tp ) < 0, which amounts to v and Tp lying in the same time cone if v is timelike.Recall also that the notion of future-oriented or past-oriented descend to curves.
From now on, we will assume that all the Lorentz manifolds we consider aretime-oriented. Moreover, we will assume that the manifold is oriented and con-nected (this ensures that we can define a global volume forme and the existence ofa complete Riemannian metric).
Definition 10.1. Let p, q ∈ M. We say that p << q if there is a future pointing timelikecurve in M from p to q and p < q if there is a future pointing causal curve in M fromp to q. Finally, p ≤ q means p = q or p < q.
Definition 10.2. For A ⊂ M, we define
I+(A) = p ∈ M : ∃q ∈ A /q << p,
J+(A) = p ∈ M : ∃q ∈ A /q < p,
I−(A) = p ∈ M : ∃q ∈ A /p << q,
J−(A) = p ∈ M : ∃q ∈ A /p < q.
The set I+(A) is called the chronological future of A, J+(A) is the causal futureand similarly for I− and J−, replacing future by past.
In the above definition, we require that the curves to be piecewise smooth. Notethat in view of the previous section, if there is piecewise smooth causal curve from pto q that is not null pregeodesic, then it can be deformed to be a piecewise smoothtimelike curve joining p to q . On the other hand, a piecewise smooth timelike curvefrom p to q can be smoothed out as follows.
Lemma 10.1. If there exist smooth timelike, future-oriented, curves from p to q andfrom q to r , then there exists a smooth timelike future oriented curve from p to r .
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Proof. Let p, q,r be as in the statement of the lemma. Let α be a smooth timelikecurve from p to q andβbe a smooth timelike curve from q to r . The curveα followedby β is not necessarily smooth at q , but we claim that we can smooth it out and stillmaintain its timelike character. First, note that this is a local problem at q , and sincethe condition of being timelike is an open condition, it follows by continuity thatit is sufficient to prove it in the case of the Minkowski metric. Let us thus assumethat η = g . Without loss of generality, we can consider coordinates (t ,r,ω) on Rn+1
such that q = (0, ..,0) and we can also assume that α : [−1,0] → M is given in localcoordinates by (s,0, ..,0)+O(s2) and that β : [0,1] → M with β(0) = q . Again, sincethis is a local problem near s = 0, we can neglect the O(s2) and we shall do so in thefollowing.
From the timelike condition, we have βt > |βi |. In particular, τ := t (β(1)) > R :=r (β(1)). We consider a curve of the form γ(s) = (
s,R f (s),ω(q))
defined on [0,τ] withf ≥ 0. The lemma then follows since we can choose f ≥ 0 such that f (x) = 0 forx ≤ 0, f (τ) = 1 and | f ′| < 1
R . For instance, rescaling in τ means we are looking fora function f defined on [0,1], such that f (x) = 0 for x ≤ 0, f (1) = 1 and | f ′| < τ
R .Draw on a picture the straight line t = x, and then draw the straight line of slope τ
Rpassing by (1,1). Then any convex function lying between the two straightlines andvanishing for x ≤ 0 will do the job.
We have easily from the definitions and Lemma 9.3.
For an open set U , we denote by I+(A,U ) the chronological future of A ⊂ U in theopen submanifold U ⊂ M . In other words, q ∈ I+(A,U ) if and only if there existsr ∈ A and a future oriented timelike curve with values in U joining r to q . In partic-ular, q ∈U , so that I+(A,U ) ⊂ I+(A)∩U .
Exercise 10.1. Give an example for which the above inclusion is strict.
Lemma 10.3 (Topological properties of causal sets in convex regions). Let C ⊂ M beopen and convex. Then,
1. For p 6= q ∈ C , q ∈ J+(p,C ) if and only if the unique geodesic from p to q is afuture oriented causal curve (similar for I+ and timelike).
2. I+(p,C ) is open in C (hence in M).
3. J+(p,C ) is the closure in C of I+(p,C ).
4. The relation ≤ is closed on C , i.e if pn → p and qn → q with all points in C thenqn ∈ J+(pn ,C ) for all n implies that q ∈ J+(p,C ).
Proof. Let p 6= q ∈C .Assume first that q ∈ I+(p,C ). Let then α be a future-oriented timelike curve
joining p to q . Since C is a normal neighborhood of p, we have α= expp β for somecurveβ : [0,1] → Tp M withβ(0) = 0. From Lemma 8.13 and the fact thatα is timelike,
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β stays in a single timecone on ]0,1]. This implies that −→pq = β(1) is timelike andfuture-oriented. Thus, the unique geodesic from p to q is a future oriented timelikecurve.
If now q ∈ J+(p,C ) and γ is a causal curve joining p to q , then either it can bedeformed in C to a timelike curve joining p to q and we run the first argument, or itis null (pre)geodesic, in which case we are also done. This proves the first point.
For the second point, let q ∈ I+(p,C ). From the first point, the unique geodesicfrom p to q is timelike. Thus g (−→pq ,−→pq) < 0. The result then follows since all theoperations here are continuous, using Lemma 8.6.
Similarly, the third and fourth points follow from the first and the continuity ofthe map (p, q) →−→pq .
Recall the definition of a continuously extendible curve, Definition 8.6.We now prove
Lemma 10.4. A causal curve contained in a compact subset K ⊂C convex is contin-uously extendible.
Proof. Let α be causal curve defined on [0,B) whose image is included K . Since Kis compact, there exists a sequence (si ) such that α(si ) → q and si → B as i →+∞.We must show that every such sequence gives the same limit. Let ti be anothersequence, such that α(ti ) → p. By the previous lemma and the fact that α is causalit follows that q ∈ J+(p,C ) and p ∈ J+(q,C ). Now if p 6= q , then from the previouslemma, −→pq is would be both future pointing and past pointing, a contradiction.
We now prove
Lemma 10.5. The relation p << q is open, that is if p << q there exists neighborhoodU of p, and V of q such that for all p ′ ∈U , q ′ ∈ V , p ′ << q ′.
Proof. Let σ be a timelike curve from p to q . If C is a convex neighborhood of q ,let q− be a point of C on σ before q . Dually, let p+ be a point of σ between p andq− and contained in a convex neighborhood C ′ of p. By the lemma, I+(q−,C ) andI−(p+,C ) are open in M and have the required property.
Corollary 10.1. For any subset A ⊂ M, I+(A) and I−(A) are open.
Proof. Follows directly from the previous lemma. (Let p ∈ A and q ∈ I+(A). Then, bythe previous lemma, there exists a neighborhood V of q including in I+(A). )
In Minkowski space, for any set A, J+(A) is the closure of I+(A), but it is easy toconstruct spacetimes where this is no longer the case.
Example: In 1+1 Minkowski space, fix a point p and consider J+(p). The bound-ary of J+(p) consists of two null curves starting at p. Remove a point on these curvesand consider the resulting spacetime to get an example where the closure of I+(p)is much larger than J+(p).
In general, we only have
Lemma 10.6. For any subset A,
1. int(
J+(A))= I+(A).
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2. J+(A) ⊂ I+(A), with equality if and only if J+(A) is closed.
Proof. I+(A) is open and contained in J+(A), hence is contained in its interior. If q ∈int
(J+(A)
), then there exists a convex neighborhood of q , C ⊂ J+(A). Take a point
r ∈C , such that r ∈ I−(q,C ). Since r ∈C , r ∈ J+(A). Hence, q ∈ I+(J+(A)) ⊂ I+(A).Since I+(A) ⊂ J+(A), we have I+(A) ⊂ J+(A). Hence, if J+(A) = I+(A), then J+(A)
is closed and if J+(A) is closed then, I+(A) ⊂ J+(A) = J+(A). Thus, the equality caseis clear provided that the inclusion assertion in (2) holds.
It suffices to just prove (2) for one point, i.e. we will prove J+(p) ∈ I+(p). Letq ∈ J+(p). If q = p, then clearly q ∈ I+(p). Otherwise, let α be a causal curve from pto q . Let C be a convex neighborhood of q and let q− be a point of α ∈ J−(q,C ). Wehave q ∈ J+(q−,C ) and by Lemma 10.6,
J+(q−,C ) = I+(q−,C ).
The result then follows since I+(q−,C ) ⊂ I+(J+(p)) ⊂ I+(p).
10.2 Various limits for sequences of curves
We are interested in the construction of limits of a sequence of causal curves. Sincewe lack good compactness properties on such sets of curves, the limits can be rough.There are several approaches to this problem:
1. Quasi-limits: this is the notion used in [O’N83]. It has the advantage that itdoes not require any knowledge of weak compactness or the Arzela-Ascoli the-orem, but the definition is quite cumbersome (see next section).
2. Limits from the C 0 topology on curves.
3. Limit curves defined using accumulation points.
We present the quasi-limits in the next section. We will use it later in order tofollow O’Neill but we also present briefly limit curves for the interested reader as in[BEE96]. The connection between the C 0 topology on curves and the limit curves isexplained in [BEE96].
10.2.1 Quasi-limits
Definition 10.3. Let (αn) be a sequence of future oriented causal curves in M and letR be a convex covering (cf Definition 8.3) of M. A limit sequence for (αn) (relative toR) is a finite or infinite sequence of points (pi ), which are causally related p0 < p1 < ...in M and such that
L1a . For each i , there exists are subsequences 0αn := αφ0(n),1αn :=0 αφ1(n),
2αn :=1
αφ2(n), .., iαn :=i−1 αφi (n) such that there exists, for each n, a set of i +1 points
along the curve iαn , denoted iαn(sn,0), iαn(sn,1), iαn(sn,2),.., iαn(sn,i ), withsn,0 < sn,1 < sn,2 < sn,3.. such that for all j ≤ i , as n →+∞,
iαn(sn, j ) → p j .
L1b . For each j < i , the points p j , p j+1 and for all n, the segments of iαn restrictedto [sn, j , sn, j+1] are all contained in a single convex neighborhood C j of the cov-ering R.
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L2. If the pi are infinite, it is nonconvergent. If they are finite, say p1 < p2 < .. < pk ,then they contains more than one point (k > 1) and there exists no pk+1 suchthat p1 < p2 < .. < pk < pk+1 and a further subsequence of kαn satisfies L1 forthe extended sequence.
Proposition 10.1. Let (αn) be a sequence of future oriented causal curves such thatαn(0) → p ∈ M and (αn) does not converge to just p (i.e. there exists a neighborhood ofp that contains only finitely many of the αn
34). Then relative to any convex coveringR, (αn) has a limit sequence starting at p.
Proof. i) Construction of the pi .Since M is paracompact35 (since every topological manifold is paracompact cf Prop.2.3 in Paulin’s differential geometry course), it has a locally finite covering R ′ by opensets B such that each B is compact and contained in some member of R. By thehypotheses on (αn), we can arrange for R ′ to be such that it contains B0 such thatinfinitely many αn start in B0 and leave B0. Relabel these curves as 1αn and for each1αn , let sn be such that 1αn(sn) is the first point in bdB0. Passing to a further subse-quence, we can assume that (1αn(sn)) converges to a point p1 ∈ bdB1. Now chooseB1 ∈ R ′ containing p1. If infinitely many 1αns leave B1, we obtain as before a subse-quence 2αn whose first departure points from B1 converges to a point p2 in bdB1.Repeat this step as many times as possible with the additional rule on subsequentchoices of Bi , if there is more than one candidate element of R ′ containing pi , pickone that has been used fewest times before.
ii) Basic properties and L1.Clearly, the first two conditions of the definition of limit sequences hold (with Ci
any element of R that contains Bi . Since the relation ≤ is closed on Ci , it followsthat pi+1 ≥ pi . By construction, pi+1 6= pi , hence pi+1 > pi .
iii) Checking L2.If the sequence (pi ) is infinite, we must show that it is non-convergent. Assume thatit converges to some q . Let B ∈ R ′ such that q ∈ B , then for all sufficiently large i ,pi ∈ B . Since B is compact and R ′ is locally finite, only finitely many number of R ′meet B . Hence, some must have been chosen for Bi for infinitely many i . But thisviolates the additional rule, for B itself was a candidate infinitely many times, butwas chosen only finitely many times (since it is open and contains all pi but a finitenumber, only a finite number can be in bd B).
Finally, suppose that the sequence of the pi is finite. Since the constructionabove cannot continue, there exists a k such that only a finite number of the kαn
leave Bk . Let (αm) be those trapped in Bk . By Lemma 10.4 (since the closure of Bk iscompact) they are extendible. Assume thus that there are defined on closed interval[0,bm] (by a change of parametrization and continuous extension, we can alwaysassume that). By compactness, we can take a further subsequence and assume thatαm(bm) converges to some q . If q = pk , then the sequence p0 < .. < pk cannot beextended to still satisfies the first two limit sequence conditions (hence it is a limitsequence). If q 6= pk , then q > pk and posing pk+1 = q , we have a limit sequence.
34The case where (αn ) just converges to p can happen by taking just one curve passing through p andshrinking the interval to 0
35Recall that this means that M is Hausdorff (séparé in French) and every open cover admits a subcoverthat is locally finite.
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Let now (pi ) be a limit sequence for (αn) and letλi be the (future causal) geodesicfrom pi to pi+1 in a convex set Ci as in L1. Assembling these segments for all i givesa broken geodesic λ called a quasi-limit of (αn) with vertices pi . λ is a future bro-ken causal geodesic that starts at p. If the (pi ) are infinite, then by L2, λ is future-inextendible. In the finite case, the curve runs from p to some pk . When all thecurves (αn) are future inextendible, we have
Lemma 10.7. A quasi-limitλ of future inextendible curves (αn) is future inextendible.
Proof. Let pi denotes the vertices of λ. If the pi are infinite, then λ is inextendible,since they must be non-convergent. On the other hand, the limit sequence cannotbe finite. Indeed, assume that the limit sequence is given by p0 < .. < pk . Let Cbe a convex neighborhood containing pk and let Bk ⊂C open and whose closure iscompact. Let kαn be a subsequence satisfying L1. Now by hypothesis, all the curvesare inextendible, thus by Lemma 10.4, they must leave Bk . But then we can pass toa subsequence to get an extra point in BdBk , contradiction.
10.2.2 Limit curves
So far we have defined the notion of causal curves for smooth curves and piecewisesmooth curves. The limit operation we will define below will in general not preservethese level of regularity so we need to introduce a broader class of causal curves.
Definition 10.4. A continuous curve γ : I → M is said to be a future directed causalcurve if for each t0 ∈ I , there exists ε > 0 such that (t0 − ε, t0 + ε) ⊂ I , and a convexneighborhood U of γ(t0−ε,t0+ε) such that given any t1, t2 satisfying t0 − ε < t1 < t2 <t0 +ε, there is a smooth future directed causal curve in U from γ(t1) to γ(t2).
Exercise 10.2. 1. Let M =S1 ×S1 be the 2-torus. Let (θ1,θ2) be standard coordi-nates on M and consider the metric
g =−dθ21 +dθ2
2 .
Prove that for any p, q ∈ M, p << q. (In the above definition, if we remove thereference to the convex neighborhood U , then every continuous curve would becausal and of course, we do not want that.)
2. Prove that piecewise smooth causal curves are causal in the sense of Definition10.4.
3. Prove that if γ : I → M is a continuous causal curve, then for any s0, s1 ∈ I ,s0 < s1, there exists a piecewise smooth future directed causal curve λ joiningγ(s0) to γ(s1).
We can now define the notion of limit curves.
Definition 10.5. A continuous curve γ : I → M is a limit curve of the sequence γn isthere is a subsequence γm such that for all p ∈ γ(I ), each neighborhood of p inter-sects all but a finite number of curves of the subsequence γm. The subsequence γmis said to distinguish the limit curve γ.
Consider normal coordinates and a normal neighborhood U around some pointp. Let t = x0 be the first coordinate, so that g t t (p) =−1. For any K > 0, consider theauxiliary Lorentzian metric
g0 =−K 2d t 2 +n∑
i=1(d xi )2.
86
Shrinking the normal neighborhood U is necessary and taking K large enough, wemay assume that the null cone of g is small than the light cone of g0, i.e. causalvectors for g are always causal for g0. Let now γ be a continuous curve such thatγ(0) = p. Let s0 < 0 < s1 be such that γ[s0,s1] ⊂ U . Then t (γ(s0)) < t (γ(s1)) since t isstrictly increasing along future directed causal curves and since we can join γ(s0) toγ(s1) by a future directed piecewise smooth causal curve. It follows that t is strictlyincreasing along γ and in particular invertible (with continuous inverse !) and wecan use it to change the parametrization of γ. Assume then that γ is parametrizedby t . Since γ is causal for g , it must be causal for g0. The causal cones of g0 are thoseof a flat spacetimes and this implies that for any t1, t2,
||γ(t1)−γ(t2)|| ≤ K ′|t1 − t2|,
where ||γ(t1)−γ(t2)|| = ||xi g amma(t1)− xi γ(t2)|| only measures the spatial partof γ. (Assume for instance that t2 = 0, then this simply says that γ(t1) lies in thefuture cone emanating from γ(t2). ) It follows that in this coordinates and with thisparametrization, γ is actually Lipschitz (in the sense that all its components are atleast Lipschitz functions of t ). Since Lipschitz functions are differentiable almosteverywhere, we can in particular define almost everywhere its tangent vector.
We now introduce an (auxiliary) complete Riemannian metric h on M . Fromthe above, for every causal curve γ defined on a compact interval, we can define itslength Lh(γ) by the usual formula, execpt that γ′ is defined only almost everwherealong the curve. This length is actually finite, as we can use the above bounds oneach time interval, after splitting the curve in sufficiently small pieces so that oneach one we have access to bounds as derived above.
Finally, an inextendible causal curve must have unbounded length (cf Hopf-Rinow). Using Arzela’s Theorem, we can then prove
Proposition 10.2. Let γn be a sequence of inextendible causal curves admittingsome p ∈ M as accumulation point. Then, there is a nonspacelike limit curve γ ofthe sequence γn such that p ∈ γ and γ is inextendible.
Proof. (sketch) Let h be an auxiliary complete Riemannian metric on M as aboveand parametrize each γn by arc length with respect to h and such that, after takinga subsequence if necessary, γn(0) → p, n →+∞. Then each γn is defined on R and
dh(γm(t1),γm(t2)) ≤ |t1 − t2|, t1, t2 ∈R.
It follows that the sequence is equicontinuous. Moreover, for any fixed t0, the se-quence restricted to [−t0, t0] is uniformly bounded. By Arzela’s theorem (exercise:state the appropriate version of this theorem), there exists a subsequence converg-ing uniformly on compact set to a continuous curve γ. Morever, we have
dh(γ(t1),γ(t2)) ≤ |t1 − t2|, t1, t2 ∈R.
It remains to prove that γ is causal and inextendible.To prove thatγ is causal, fix t1 and let U be convex withγ(t1) ∈U . Chooseδ> 0 so
that the open ball for h, Bh(γ(t1),δ) is in U . For t2 ∈ (t1, t1+δ), we have γn[t1, t2] ⊂Uand using that the relation p ≤ q is closed in U (cf Lemma 10.6), we obtain that γ(t1)can be joined by a future directed causal curve to γ(t2).
Assume finally that γ is not future inextendible and thus that γ(t ) → q as t →+∞. Take U a convex neighborhood of q with compact closure and such that in U ,
87
(xα) are coordinates with x0 strictly increasing along timelike curve (as in the abovediscussion). There exists a t1 such that γ restricted to [t1,+∞) lies U .
On the other hand, using the length function Lh and bounds as above, one canprove (exercise) that a causal curve with initial x0 = x0(γ(t1) and ending at a pointwith x0 = x0(q) has length uniformly bounded by some L. On the other hand, fork sufficiently large the component x0 of γk restricted to [t1 + 1, t1 +L + 2] must liewithin [x0(γ(t1), x0(q)] by uniform convergence, continuity and the fact that x0 isstrictly increasing along γ. In view of the arc length parametrization, the length ofγk restricted to [t1 +1, t1 +L+2] is L+1 which is a contradiction.
10.3 Causality conditons
First, an easy remark
Lemma 10.8. A compact Lorentzian manifold admits a closed timelike curve.
Proof. By compactness, the open covering I+(p) admits a finite subcover I+(p1), .., I+(pk ).If p1 ∈ I+(pi ), for i 6= 1, then I+(p1) ⊂ I+(I+(pi ) = I+(pi ). In that case (note that thenk > 1), we can remove I+(p1) and still obtain a open cover. Wlog, assume thus thatp1 is not included into any of the other I+(pi ). Then, p1 ∈ I+(p1), i.e. there exists aclosed timelike curve at p.
Closed timelike curves are bad: from the physics point of view, and from evolu-tion problems point of view. We want to exclude them. We want in fact a slightlystronger criterion.
Definition 10.6. The strong causality condition is said to hold at p ∈ M providedthat given any neighborhood U of p, there exists a neighborhood V ⊂ U of p suchthat every causal curve segment with endpoints in V lies entirely in U .
Exercise 10.3. 1. Prove that, in the above definition, one can use piecewise smoothcausal curves or continuous causal curves equivalently.
2. Prove that if there exists a closed timelike curve passing through p, then M isnot strongly causal at p.
Lemma 10.9. Suppose that the strong causality condition holds on a compact K ⊂ M.If α is a future-inextendible causal curve that starts in K , then α eventually leaves Knever to return; that is there is an s > 0 such that α(t ) ∉ K , for all t ≥ s.
Proof. Assume that α persistently returns to K . Assume that α is defined on [0,B).Then, there exists a sequence si → B , with α(si ) ∈ K . Taking a subsequence, we canassume that α(si ) converges to some p ∈ K . Since α has no future endpoint, theremust exists another sequence t j → B , such that α(t j ) does not converge to p. Tak-ing yet another subsequence, we can assume that there exists a neighborhood U ofp such that α(t j ) ∉U for all j . Since s j and t j both converges to B , they have sub-sequences that alternates s1 < t1 < s2 < t2... The curves α|[sk ,sk+1 ] are almost closedat p: Given any neighborhood V ⊂ U of p, for k large enough the ends α(sk ) andα(sk+1) are contained in V and yet the curve escape U since α(tk ) ∉U .
Lemma 10.10. Suppose the strong causality condition holds on a compact subset K .Let (αn) be a sequence of future oriented causal curve segments in K such thatαn(0) →p and αn(1) → q 6= p. Then, there exists a future oriented causal broken geodesic λfrom p to q and a subsequence αm whose length converges with the limit satisfyinglimm→+∞ L(αm) ≤ L(λ).
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Proof. From Proposition 10.1, the αn admits a limit sequence pi starting at p. If pi
is infinite, the corresponding quasi-limit λ is future inextendible and thus, by theabove lemma, must leave K . In particular, some of the αn must leave K , a contra-diction. Thus the limit sequence is finite, starts at p and ends at q (again applyingL2). The quasi-limit with these vertices is a broken geodesic from p to q . Thanksto L1b, we can consider a subsegment in some convex set Ci . On each convex set,the length separating two points p < q is bounded above by the length of the corre-sponding geodesic. Thus,
L(αm|[sm,i ,sm,i+1]) ≤ | ~pmi pm,i+1|,
where pmi = α(smi ). Moreover, we know that the length (in fact ~pq itself) betweentwo points p < q in a convex set depends continuously on p and q . Thus, | ~pmi pm,i+1|, |→~pi , pi+1|.Suming on all the segments and taking the limit m → +∞ then gives the
lemma.
10.4 Achronal sets
Definition 10.7. A ⊂ M is said to be achronal if there exists no p, q ∈ A such thatp << q and acausal if there exists no p, q such that p < q. Given A achronal, D+(A)is by definition, the set of all p ∈ M such that every past inextendible causal curvethrough p meets A. Similarly, we define D−(A) as the set of all p ∈ M such that everyfuture inextendible causal curve through p meets A.
Definition 10.8. The edge (marge in French, not to be confused with the boundary intopological sense) of an achronal set A consists of all points p in the closure of A suchthat every neighborhood U of p contains a timelike curve from I−(p,U ) to I+(p,U )that does not meet A.
Exercise 10.4. Let A be the set [0,1]×t = 0 in 2-dimensional Minkowski space. Whatis the edge of A ?
Recall the definition of a topological hypersurface of M .
Definition 10.9. S is a topological hypersurface of M if for all p ∈ S, there exists aneighborhood U of p in M and a homeomorphism φ : U → Rn such that φ(U ∩S) =φ(U )∩ΠwhereΠ is a hyperplane in Rn .
Remark 10.1. The point is that topological hypersurface are rougher than differen-tiable submanifold. In particular, they do not necessarily admits tangent planes.
Lemma 10.11. Let p ∈ M and U a neighborhood of p. Then, there exists a coordinatechat (N ,ξ) of p such that
1. ξ(N ) has the form (a −ε,b +ε)×N ⊂ R1 ×Rn−1 with a < b, ε> 0.
2. For any (xi ) ∈ N , the curve s → ξ−1(s, xi ) is timelike.
3. The slice x0 = a of N is in I−(p,U ), the slice x0 = b of N is in I+(p,U ).
Proof. Consider a normal neighborhood U of p with U ⊂ U . Let ξ be a normal co-ordinate system at p with g (∂x0 ,∂x0 )(p) =−1 and ∂x0 future pointing. Consider theregion N such that, |x0| ≤ δ and |(xi )2| ≤ 1/8δ2. Let V be defined by V = exp−1
p (N )and let Vρ be the subset of vectors which are timelike for the auxiliary Minkowski
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metric η1/4 :=−1/4d x0 ⊗d x0 +d xi ⊗d xi . Now, let β be a curve in Vρ which is time-like for η1/4. Let yα(t ) be the coordinates of β
β(t ) = yα(t )[∂xα ]|p
and let γ = expp β. Then the normal coordinates of γ(t ) are also given by yα(t ) byconstruction.
Moreover, by construction, the coordinates of γ′ are given by (yα)′ and
−1/4((y0)′)2 + ((y i )′)2 < 0.
Since at p, the metric in normal coordinates coincides with the Minkowski met-ric, by taylor expansion, we have that,
gαβ(x) = ηαβ+O(|x|)
and hence, for δ small enough,
|gαβ(x)−ηαβ| ≤Cδ,
for some C .This implies that
g (α′,α′) = η(α′,α′)+O(δ)(|y ′|2),
where |y ′|2 = ((y i )′)2 + ((y0)′)2 ≤ 5/4((y0)′)2. Since η(α′,α′) < −3/4((y0)′)2, we havethat for δ small enough, g (α′,α′) < 0.
In particular, expp maps each xi = const curve to a timelike curve in N . We
claim that the points of normal coordinates (−δ, xi ) and (δ, xi ) lies in respectivelyI−(p,U ) and I+(p,U ). Indeed, let v = δ[∂x0 ]p + xi [∂xi ]p and consider the curveβ : t → t v . Then, β′ = v and η1/4(v, v) < 0. Thus, β is timelike for the auxiliary metricand γ = expp β is a future timelike (for g ) geodesic from p to the point of normal
coordinate (δ, xi ). Finally, for ε> 0 sufficiently small, we have again that the points ofnormal coordinates (−δ+ε, xi ), (δ−ε, xi ) lies in respectively I−(p,U ) and I+(p,U ).
Proposition 10.3. An achronal set A is a topological hypersurface if and only if Acontains no edge point (i.e. A and edge A are disjoints).
Proof. First, let A be a topological hypersurface. Let p ∈ A and let U be an neigh-borhood of p as in Definition 10.9. Without loss of generality, we can assume that U
is connected and that U \A has two connected components. (Consider φ(p) in Rn .There exists a ball B centered at p such that B ⊂ φ(U ). Then φ−1(B) is a neighbor-hood of p ∈ M that works). Since A is achronal, the open sets I−(p,U ) and I+(p,U )are disjoints. For otherwise, there exists a closed (piecewise) smooth timelike curvethrough p ∈ A, which contradicts that A is achronal. For similar reasons, they donot meet A. Moreover, I−(p,U ) and I+(p,U ) are connected, since otherwise U \ Ahas at least three connected components. Any timelike curve through p meets bothsets, hence they are contained in different components of U \ A. Thus every timelikecurve from I−(p,U ) to I+(p,U ) must go through some point of A. Thus, p is not inthe edge of A.
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Now suppose that A and edge A are disjoints. Let p ∈ A. Since p ∉ edge(A),there exists a neighborhood U of p, such that every timelike curve from I−(p,U ) toI+(p,U ) intersects A.
From the previous lemma, we consider a neighborhood N of p such that
1. ξ(N ) has the form (a −ε,b +ε)×N ⊂ R1 ×Rn−1.
2. The slice x0 = a of N is in I−(p,U ), the slice x0 = b of N is in I+(p,U ).
If y ∈ N ⊂Rn−1, the x0 coordinate curve
s → ξ−1(s, y)
is timelike by construction, and must then meet A. Since A is achronal, the meet-ing point is unique. Let h(y) be its x0 coordinate. It suffices now to show that thefunction
h : N → (a,b)
is continuous, for thenφ= (x0−h(x1, .., xn−1), x1, .., xn−1) is a homeomorphism fromN onto its image (easy to check, just write explicity φ−1 ) that carries A ∩N to theslice u0 = 0 of φ(N ) ⊂ Rn .
Let yn be a sequence that converges in N to y . Asssume that h(yn) does notconverge to h(y). Then, there must exists (since the h values are bounded) a sub-sequence yψ(n) converging to some r 6= h(y). Now since r 6= h(y), either r > h(y),in which case one can reach ξ−1(y,r ) from ξ−1(y,h(y)) by following ∂x0 , a timelikevector field forward, or r < h(y) and then we follow ∂x0 backwards. Thus, ξ−1(y,r )is in the set I−(q,N )∪ I+(q,N ), where q = ξ−1(y,h(y)). Since this set is open, itfollows that there exists some n such that ξ−1(yn ,h(yn)) ∈ I−(q,N )∪ I+(q,N ) forsome n. But these belong to A so they cannot be in I−(q,N )∩ I+(q,N ) since A isachronal.
Remark 10.2. One can in fact show that such an A is a Lipshitz submanifold, i.e. thefunction h constructed above satisfies a Lipshitz condition, cf Penrose [Pen72].
Corollary 10.2. An achronal set A is a closed topological hypersurface if and only ifedge A is empty.
Proof. Assume that A achronal is a topological hypersurface. Then A and edge A aredisjoints. But edge A is included into the closure of A, so if A is closed then edge Amust be empty.
Reciprocally, suppose edge A is empty. Then A is topological hypersurface. More-over, if A is achronal so if A (because the relation << is open). Thus, if q ∈ A \ A, thenno timelike curve through q can ever meet A (otherwise, since A ⊂ A, that wouldcontradicts that A is achronal), which implies that q is in edge(A) which we assumedwas empty.
Definition 10.10. A subset F of M is a future set (respectively past set) providedI+(F ) ⊂ F (respectively I−(F ) ⊂ F ).
Example: For any set B , J+(B), I+(B) are future sets.
Exercise 10.5. If F is a future set, then M \ F is a past set.
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Corollary 10.3. The topological boundary of a future set is a closed achronal topo-logical hypersurface if non-empty.
Proof. 1. ∂F is achronal: Let F be a future set and p ∈ ∂F . If q ∈ I+(p), then I−(q) is aneighborhood of p and hence contains a point of F . Thus, q ∈ I+(F ) ⊂ F , using thatF is a future set. This proves that I+(p) ⊂ F and similarly, one has I−(p) ⊂ M \ F . Inparticular, I+(∂F ) and I−(∂F ) are disjoint and hence ∂F is achronal.
2. The closed set ∂F has no edge points: Indeed, I+(p) ⊂ intF and I−(p) ⊂ extFfor p ∈ ∂F . (Let U be a neighborhood of p and γ be a timelike curve from I−(p,U )to I+(p,U ), then since I−(p,U ) and I+(p,U ) are contained in intF and extF , γ mustintersect the boundary of F ). The conclusion then follows from the previous corol-lary.
Remark 10.3. This last corollary has in fact plenty of applications in the study ofpossible singularities for non-linear wave equations. For instance, consider a waveequation of the form
äu = F (u,u′)
where F is some non-linearity and ä is the usual wave operator. We imagine thatwe have initial data given at t = 0 and that the problem is well posed at locally, sothat we can construct for any data a local solution. Now, using the domain of thedependence/finite speed of propagation, we can consider the largest domain on whichthe solution stays regular (see Alinhac’s book [Ali09] for more on how to define thisproperly). That domain will be a future set, and hence its boundary (where somethingbad has to happen to our solution, otherwise we would continue) must be at least aclosed achronal topological hypersurface.
10.5 Cauchy hypersurfaces
Definition 10.11. A Cauchy hypersurface in M is a subset S that is met exactly onceby every inextendible timelike curve in M.
Exercise 10.6. 1. Prove that any t = const hypersurface in Minkowski space is aCauchy hypersurface.
2. In 2d Minkowski space, for any ρ > 0, prove that the hyperboloid
(t , x) : t 2 −x2 = ρ, t > 0
is however not a Cauchy hypersurface.
We will need the following technical lemma.
Lemma 10.12. Let α be a past-inextendible causal curve starting at p that does notmeet a closed set C . Then, if p0 ∈ I+(p, M \ C ), there is a past inextendible timelikecurve through p0 that does not meet C .
Remark 10.4. Note that the original curve is only causal while the final curve is time-like. On the other hand, the point p0 is in I+(p, M \ C ) so it can be joined to p by atimelike curve.
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Proof. Since α is past inextendible, we may assume that α has domain [1,+∞) andthat the sequence α(n) does not converge.
Since p0 ∈ I+(p, M \ C ), we can join p0 to α(1) = p by a timelike curve that doesnot meet C . Then the combined curve, denoted γ, gives a causal curve starting atγ(0) = p0 which is inextendible and has values in M \ C . Moreover, γ is not a nullpregeodesic since it is initially timelike.
We claim that we can deform it, to be timelike everywhere and still avoidingC . We will denote β the deformed curve. Indeed, consider the points p1 = α(1),p2 =α(2), .., pn =α(n). First, we can deform γ on [0,2] to be a timelike curve keep-ing the ends p0 and p2 fixed and being arbitrary close to γ. Since γ([0,2]) is compactand C is closed, if the deformation is small enough then the deformed curve β′ doesnot meet C . (For instance, everything can be quantified using a metric d on M). Wethen define β on [0,1] to be the curve just obtained. To construct β on [0,2], we con-sider the curve β′ constructing previously on [1,2] (and agreeing with β on [0,1] byconstruction ) and join it toα from [2,3]. This is a causal curve fromβ(1) to p3 whichis initially timelike. Thus, it can be deformed so that the deformed cuve β′′ is time-like everywhere and avoids C . Extending β on [0,1] by β′′ on [1,2] we obtain β on[0,2]. Repeating the process, we construct β on [0,∞), which is timelike everywhere,avoids C and moreover β(n) can be taking arbitrarily close to α(n) depending on n.For instance, we can arrange d(β(n),α(n)) < 1/n, for an arbitrary metric d on M . Itfollows that β is inextendible.
We have
Proposition 10.4. A Cauchy surface is a closed achronal topological hypersurfaceand is met by every inextendible causal curve.
Remark 10.5. Note that in Definition 10.11, a Cauchy surface was only assumed tobe met by timelike curves.
Proof. Let p ∈ M . Let γ be a maximally defined (thus inextendible) future orientedtimelike geodesic γ defined in a neighborhood of 0 and such that γ(0) = p. Then, bydefinition of a Cauchy hypersurface, γ intersect S once and only once, i.e there existsa unique t0 such that γ(t0) ∈ S. Now, depending on whether t0 > 0, = 0, or < 0, wehave that p ∈ I+(S), S, or I−(S). Thus, M is the disjoint union of I+(S), S and I−(S).Moreover, for any q ∈ S, any future oriented timelike curve γ such that γ(0) = q issuch that γ(t ) ∈ I+(S) for t > 0 and γ(t ) ∈ I−(S) for t < 0 and in particular, S ⊂ ∂I±(S)and in fact S is the common boundary of the sets I±(S) (note that ∂I+(S)∩I−(S) =6= 0since I−(S) is open and disjoint from I+(S)).
From Corollary 10.3, it follows that S is a closed achronal topological hypersur-face. It remains to show that S is met, not just by every inextendible timelike curve,but by every inextendible causal curve α.
Assume that α is an inextendible causal curve that does not meet S. Wlog, as-sume that α(0) ∈ I+(S). Then, by the previous lemma, there is a past inextensibletimelike curve β starting in I+(S) that does not meet S. Since any future-pointingtimelike curve starting at β(0) must remain in I+(S), thus adjoining any such curveto β gives an inextendible timelike curve that avoids S, i.e. a contradiction.
Proposition 10.5. Let S be a smooth spacelike Cauchy hypersurface. Then M is dif-feomorphic to R×S. Futhermore, if S′ is another smooth Cauchy hypersurface, thenS′ and S are diffeomorphic.
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Proof. Let T be a globally defined timelike vector field whose flow is defined on R×M , i.e. T is complete, cf. Proposition 8.5. Its restriction to R×S gives a smooth map
f :=φ|R×S :R×S → M .
First, we claim that f is injective. Indeed, since if φ(t1, x1) = φ(t2, x2), then byuniqueness and definition of the flow (or the semi-group property of t →φ(t , .) ), wemust have x1 =φ(t2 − t1, x2). Since S is achronal and T timelike, we must then havet2 = t1 and x2 = x1.
Let now (t0, x0) ∈ R× S. We claim that d f(t0,x0) is surjective. Indeed, let h bedefined h(p) = φ(−t0, p), then h is a diffeomorphism of M and d f(t0,x0) is surjectiveis and only if d(h f )(t0,x0) is surjective. Now by construction, for any (t , p) ∈R×S,
h f (t , p) = h(φ(t , p)) =φ(t − t0, p).
Consider any curve s → (t ,γ(s)) where γ is a curve in S, as well as any curve of theform s → (s, p) for p ∈ S, it follows that the image of d(h f )(t0, x0) contains both Tx0
and the tangent space of S. Since S is spacelike and Tx0 timelike, it follows that theimage of d(h f )(t0, x0) has dimension n and that d(g f )(t0,x0) is surjective.
Since f is a local diffeomorphism and injective, it is in fact a global diffeomor-phism onto its image. We claim that it is also surjective. Indeed, let p ∈ M and letγ be the maximal integral curve of T starting at p. Then γ is inextendible and thusintersect S. It then follows that p is in the image of f .
Let now π : R×S → S be the projection on S. If S′ is another smooth spacelikeCauchy hypersurface, then we get a map
r : S′ → S
x → r (x) =π f −1(x).
Then, r is smooth and r maps a point x ∈ S′ to the point of intersection between theintegral curve of T starting at x and S. The inverse of r can be defined as the mapthat takes y ∈ S to the point of intersection between the integral curve of T throughy and S′. Thus, S and S′ are diffeomorphic.
Recall that a Cauchy hypersurface needs not be differentiable, but is a topologi-cal hypersurface at minimum. In that latter case, we have
Proposition 10.6. Let S be a Cauchy hypersurface in M and let X be a timelike vectorfield on M (which exists, since M is time-oriented). If p ∈ M, the maximal integralcurve of X through p meets S at a unique point ρ(p). Then ρ : M → S is a continuousopen map onto S leaving S pointwise fixed. In particular S is connected (assuming Mconnected).
Proof. See O’Neill [O’N83, p.417] (the proof requires invariance of domain, it wouldbe nice to have another proof not based on it, though unlikely to be true since weare dealing with only topological manifolds here).
10.6 Global hyperbolicity
The notion of global hyperbolicity was introduced by Leray36 [Ler55] in a more gen-eral context arising in the study of general hyperbolic pdes.
36His definition was naturally slightly different, in particular outside from the realm of Lorentzian ge-ometry. Instead of the compactness of J−(p)∩ J+(q), he asked a C 0 compactness on the set causal curvesfrom p to q .
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Definition 10.12. A spacetime is said to be globally hyperbolic if it is strongly causaland if for any p, q, J−(p)∩ J+(q) is compact.
We have the following results.
Proposition 10.7. If (M , g ) is globally hyperbolic, then the causality relation ≤ isclosed, i.e. if pn → p and qn → q, with pn ≤ qn , then p ≤ q.
Remark 10.6. Recall that in Lemma , we had proved that ≤ was closed when re-stricted to a convex set.
Proof. The proof is trivial is p = q . Hence, we can assume that p 6= q . Let αn be acausal curve from pn to qn . Take p− << p and q << q+. Then, for n large enough,p− << pn ≤ qn << q+. Thus, the curveαn are all in the compact set J+(p−)∩ J−(q+).From Lemma 10.10, there exists a causal curve from p to q .
Lemma 10.13. If (M , g ) is globally hyperbolic and p < q, then there is a causal geodesicfrom p to q such that no causal curve connecting p to q has greater length.
Proof. Since p < q , the set C (p, q) of causal curves from p to q is non-empty. Letτ(p, q) = supγ∈C (p,q) L(γ). Note that a priori, τ(p, q) ∈ [0,+∞]. Let αn be causalcurves from p to q such that L(αn) → τ(p, q) as n → +∞. These curves are all inJ+(p)∩ J−(q) which is compact and strongly causal by global hyperbolicity. Hence,by Lemma 10.10, there is a causal broken geodesic λ from p to q with L(λ) = τ(p, q).Moreover, if there is a break say at some r with p ≤ q ≤ r , then take a normal neigh-borhood U at r and consider points along the curve p ′, q ′ ∈U , with p ′ ≤ r ≤ q ′. Then,from Lemma 8.14, the geodesic segment from p ′ to q ′ is timelike and has biggerlength than the corresponding segment of the original curve, a contradiction.
Lemma 10.14. Let (M , g ) be a globally hyperbolic Lorentzian manifold and let K1,K2 ⊂M be compact. Then J−(K1)∪ J+(K2) is compact.
Proof. Recall that since M is metrizable (since M is a manifold), a subset is compactiff it is sequentially compact. Let qn ∈ J−(K1)∪ J+(K2). By assumptions, there existspn,i ∈ Ki , such that pn,2 ≤ qn ≤ pn,1. Since the Ki are compact, up to subsequences,we can assume that pn,i → pi . Let ri ∈ M be such that r2 << p2 and p1 << r1. Then,r1 << qn << r2 for n large enough. Hence, qn ∈ J−(r1)∪ J+(r2) for n large enoughand since this set is compact by global hyperbolicity, up to a subsequence qn con-verges to some q . Since the relation ≤ is closed on globally hyperbolic manifold, weconclude that p2 ≤ q ≤ p1.
10.7 Cauchy developments
Definition 10.13. If A is an achronal subset of M, the future Cauchy developmentof A is the set D+(A) of all points p ∈ M such that every past-inextendible causalcurve through p meets A. In particular A ⊂ D+(A). Similarly, D−(A) is the set ofall points p ∈ M such that every future-inextendible causal curve through p meets A.The Cauchy development of A is by definition D+(A)∪D−(A).
In particular, if S is a Cauchy hypersurface, then D(S) = M .
We have easily
Lemma 10.15. 1. D+(A) ⊂ A∪ I+(A) ⊂ J+(A),
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2. D+(A) and I−(A) are disjoints,
3. D+(A)∩D−(A) = A and D+(A) \ A = D(A)∩ I+(A).
4. If α is a past-directed causal curve starting in D+(A) and leaving D+(A), thenit first leaves D+(A) through A.
Proof. We only prove the last claim, the first four follows from the definitions (andthe fact that A is achronal!). Let α= [0,b) → M be a past-directed causal curve suchthat α(0) ∈ D+(A). Let s > 0 such that α(s) ∉ D+(A). By definition of D+, there mustbe a past-inextendible causal curve β starting at α(s) that does not meet A. Butα|[0,s] +β must meet A, thus α|[0,s] must intersect A say at some t0. If there existsanother 0 < s′ < t0 such that α(s′) ∉ D+(A), then we can produce another t1 < t0
such that α(t1) ∈ A, which contradicts the achronality of A.
We will also need.
Lemma 10.16. If A is achronal and p ∈ intD(A), then every inextendible causal curvethrough p meets both I−(A) and I+(A).
Proof. Since D(A) ⊂ A ∪ I+(A)∪ I−(A), we can assume wlog that p ∈ A ∪ I+(A). Letα : [0,+∞) be an inextendible causal curve starting at p. Consider the points pi =α(i ) alongα. Let d be an auxiliary metric on M . Let r0 ∈ A be a point such that r0 >>p0 = p. By induction, we can construct, a sequence of points (ri ) in A such that,for i ≥ 1, ri−1 >> ri >> pi and d(ri , pi ) < 1/i . The curve β obtained by joining theri by timelike segment is then a past directed inextendible (the ri cannot converge)timelike curve such that the past of any point of the curve contains a point of α.Since β is past inextendible, it must intersect A and thus α ∈ I−(A).
(To get an idea why p must lie in the interior, just take A a closed horizontal in-terval in 2d Minkowski space and then take p on the boundary of D(A) in the futureof A. )
Lemma 10.17. Let A be achronal and p ∈ intD(A) \ I−(A). Then, J−(p)∩D+(A) iscompact.
Proof. Let xn be a sequence in J−(p)∩D+(A). If xn admits a subsequence converg-ing to p we are done, so we assume that it is not the case. In particular, we mayassume that xn 6= p for all n sufficiently large. Let αn be past pointing causal curvesfrom p to xn . Using Proposition 10.1, since no subsequences of xn converges to p,there is a past directed limit sequence pi for αn starting at p. If pi is infinite,then if λ is a quasi-limit, λ is past inextendible casual curve starting at p. Thus, byLemma 10.16, it must intersect I−(A), which is a contradiction. Thus, the pi are fi-nite and then some subsequence of the xn (which are the endpoints of the αn) mustconverges to the last pi , denoted here x, with x ∈ J−(p). It remains to prove thatx ∈ D+(A). Note first that x 6= I−(A), as this would imply that some xn ∈ I−(A), whichis impossible since xn ∈ D+(A) and A is achronal.
Since p ∈ intD(A) \ I−(A), there exists q ∈ D+(A)∩ I+(p). Let then σ be a pastinextendible timelike curve σ from q =σ(0) passing through x =σ(t0), at some t0 >0. (for instance σ can be constructed as in the previous lemma). Since q ∈ D+(A)and A is achronal, σ must intersect A once and only once, say at some t1. Note thatit then follows that σ|[0,t1] ⊂ D+(A). Since x 6= I−(A), it follows that we must havet0 ≤ t1 and then x ∈ D+(A).
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Theorem 10.1. If A is an achronal set, then intD(A), if non-empty, is globally hyper-bolic.
Proof. • There exists no closed causal loops: If there is a closed causal loops,by passing through the loops infinitely many times, we can construct an in-extendible causal loops. This must then intersect A infinitely many times andusing 10.16, contradicts that A is achronal.
• Assume that the strong causality causality does not holds at some p ∈ intD(A).Then, there exists a neighbordhood U ∈ intD(A) of p and future directed causalcurves segments αn : [0,1] →U such that αn(0) and αn(1) converges to p butnone of the αn are contained in U . Thus, αn has a future directed limit se-quence at pi starting at p. If the pi are finite, then it must ends at p (if it endsat some p ′ 6= p, then we can extend the limit sequence, contradiction). It thenhollows that p ∈ J+(p), i.e. there is a closed future directed causal curve from pto p, which we have already excluded. Thus, the pi are infinite and the corre-sponding quasi-limit λ is future inextendible. From Lemma 10.16, there existssome t0 such that λ(t0) ∈ I+(A) and from the achronality of A, λ(t ) ∈ I+(A),for all t ≥ t0. In particular, some point pi of the limit sequence must be-long to I+(A). Thus, there is a subsequence αm and some sm ∈ [0,1] suchthat αm(sm) ∈ I+(A) → pi . Reparametrizing if necessary, we may assume thatsm = s is independent of m. Since pi 6= 0, we can obtain another past-directedlimit sequence qi starting at p by considering the curvesαm restricted to [s,1].If the limit sequence is finite, it must end at pi , and then pi < p. Since, p < pi ,we again obtain a causal loop. Thus, the limit sequence is infinite and the cor-responding quasi-limit σ is a past-directed inextendible causal curve startingat p. By Lemma 10.16 again, there exists some t0 such that σ(t0) ∈ I−(A) andin particular, some αm restricted to [s,1] must meet I−(A). Since αm is futurepointing and has αm(s) ∈ I+(A), we again contradict the achronality of A.
• Let p 6= q , with p, q ∈ intD(A). If p = q , then, since there are no causal loops,J−(q)∩ J+(p) = p is compact. Suppose then that p < q . Let xn be a se-quence in J−(q)∩ J+(p). For any n, let αn be a future directed causal curvefrom q to p passing by xn . Let R be a covering of M by convex open sets Csuch that C is compact and contained in a convex open set. Let pk be a limitsequence relative to R starting at p.
1. If the pi are finite, they must end at some pk = q . Let (αm) = (kαm)be a subsequence as in Definition 10.3, L1a. Recall that, for any m, xm
belongs to αm . Thus, it must belongs to one of segments [sm, j , sm, j+1],where j potentially depends on m. On the other hand, since, for any m,they are only finitely many such segments (indexed by j not m !), thereexists i < k such that for infinitely many m, xm ∈ [sm,i , sm,i+1]. Thus,there exists a subsequence, still denotedαm , such that xm ∈ [sm,i , sm,i+1],for all m and hence the xm all belongs to a single convex set C of compactclosure. In particular, they admits a subsequence converging to some x,and since ≤ is closed on convex sets, we have pi ≤ x ≤ pi+1 and hencex ∈ J−(q)∩ J+(p).
2. If the pi are infinite, they the corresponding quasi-limit is a future inex-tendible causal curve starting at p. The proof is similar to the proof of thestrong causality condition as above. First, there exists a subsequenceαm
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and after a change of parametrization a fixed s such that αm(s) convergeto some point pi ∈ I+(A). Since pi 6= q , we can consider αm restrictedto [s,1] and construct a past directed limit sequence qi starting at q . Ifthe limit sequence is finite, then it must ends at limα(s) = pi and it thenfollows that p < p1 < ... < pi < ... < q1 < q is a finite limit sequence forαn , which we have already ruled out. Thus, qi is infinite and the corre-sponding quasi-limit is a past inextendible causal curve µ starting at q .µ must interstect I−(A) hence some αm restricted to [s,1] does and thiscontradicts achronality.
• This proves that J−(q)∩ J+(p) is compact, but a priori this sets may differ fromJ−(q, intD(A))∩ J+(p, intD(A)). We now prove that J−(q)∩ J+(p) ⊂ intD(A),which implies that the two above sets are the same. As before, we may assumethat p < q . It will be sufficient to consider only two cases p, q ∈ I+(A) or p ∈J−(A), q ∈ J+(A) (the other cases are similar exchanging − and +. Note alsothat I±(A) ⊂ J±(A) ).
1. p, q ∈ I+(A). Let q+ ∈ I+(q)∩D(A) ⊂ D+(A). Then, U = I+(A)∩ I−(q+)is an open set containing J−(q)∩ J+(p). Let σ be a past directed timelikecurve from q+ to y ∈U . Then, since A is achronal and y ∈ I+(A), σ doesnot meet A, thus we must have y ∈ D+(A).
2. p ∈ J−(A), q ∈ J+(A). Similarly, we can find points p− ∈ I−(A)∩D−(A)and q+ ∈ I+(q)∩D+(A) and consider the open set U = I+(p−)∩ I−(q+).Let x ∈ U , σ be a past directed timelike curve from q+ to x and τ be apast directed timelike curve from x to p−. Since A ⊂ D(A), we can alsosuppose that x 6= A. By achronality, at least one of τ and σ do not meetA, say σ. This then implies that x ∈ D+(A).
The last theorem implies in particular
Corollary 10.4. A spacetime that admits a Cauchy hypersurface is globally hyper-bolic.
Proof. This follows from M = D(S), if S is a Cauchy hypersurface and the previoustheorem.
10.8 Time and temporal functions
In the whole section, (M , g ) is a globally hyperbolic Lorentzian manifold.The aim of the section is to prove the existence of certain special, globally de-
fined, functions on globally hyperbolic Lorentzian manifold. These functions willhave the property that each of their level sets will be Cauchy hypersurfaces. More-over, we will require more and more structure on them to eventually construct asmooth function, whose level sets are smooth spacelike Cauchy hypersurfaces andsuch that its gradient is timelike and past directed. This will be our replacement forsay, the function t in Minkowski space.
First, the functions that we will construct will lack regularity. We will then modifyit locally near any level set to eventually reach our goal.
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10.8.1 Time function and Geroch’s theorem
Let M be oriented, so that we can define globally a volume form
η=√|det g |d x1 ∧ ..d xn
Recall the definition of a smooth partition of unity.
Definition 10.14. Given a open covering U = Ui of M, a smooth partition of unityfor the Ui is a collection of smooth maps φi such that
• For any i , φi : M → [0,1].
• There is a locally finite open refinement V j of U such that the support of fi isincluded in Vi for any i ,
•∑
i fi = 1.
Let (Ui ,ξi )1≤i≤+∞ be an atlas of M compatible with its orientation (so that the Ja-cobian matrices of the change of coordinates have positive determinant on overlap)and let (φi ) be a smooth partition of unity for the Ui . Finally, let mi be the integralof the φi .
Then, defines the volume form
ω=+∞∑i=1
1
mi 2iφi η.
Note that by construction,∫M ω= 1 <+∞, so that we have defined a finite measure on M .
LetΛ be the linear functional that takes smooth function of compact support onM toΛ( f ) = ∫
M f ω.By Riesz representation theorem, there exists a positive measure µ and a σ alge-
bra of subsets of M containing the Borel sets of M , such that µ is complete and suchthat for smooth functions of compact support,∫
f dµ=Λ( f ).
By construction, µ(M) = 1 (this follows from the monotone convergence theorem),µ(U ) > 0 for any open non-empty set and for any measurable set V , µ(V ) = 0 if andonly if for all i , ξi (V ∪Ui ) is of zero measure for the Lebesgue measure.
We will use this measure to prove the following theorem.
Theorem 10.2 (Geroch’s theorem). There exists a continuous, onto, function τ : M →R which is strictly increasing along any future oriented causal curve and such that ifγ is an inextendibly causal curve defined on (t−, t+), then τ(γ(t )) →±∞, as t → t±. Inparticular, for any a ∈R is an acausal Cauchy hypersurface.
Proof. The fact that if τ is such a function, then its level sets are Cauchy hypersurfacefollows from the definition of a Cauchy hypersurface. Since the monotonicity of τholds for any causal curve (not just timelike curve), then the levet sets are acausal:they can not be met more than once by causal curves.
Let µ be the measure constructed above. Let p ∈ M and let C+p = J+(p) \ I+(p).
We claim thatµ(C+
p ) = 0.
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Indeed, let q ∈ C+p . Then, there is a causal curve from p to q . Moreover, since M
is globally hyperbolic, from Lemma 10.13, there exists a causal geodesic connectingp to q of greater length and since q ∉ I+(p), it must be a null geodesic. Thus, thereexists a v ∈ Tp M which is null and such that expp (v) = q . Let N be the future directednull vectors in Tp M that lies in the domain of expp . This is an open submanifold ofTp M of dimension n−1. Moreover, expp is a smooth map from N to M and C+
p \p ⊂expp (N ). An application of Sard’s theorem then implies that µ(expp (N )) = 0 andhence C+
p has also zero measure.Similarly, with C−
p = J−(p) \ I−(p), we have µ(C−p ) = 0.
Let us define now f−(p) =µ[J−(p)](=µ[I−(p)]) and f+(p) =µ[J+(p)](=µ[I+(p)]).Since I±(p) are open and non-empty, f±(p) > 0, for all p ∈ M .
Step 1: f− and f+ are continuous.
Let p j → p.
Assume first that p j << p. Since J−(p j ) ⊂ J−(p), we know that f−(p j ) ≤ f−(p).In view of the preliminary step, for all q ∈ M ,
f−(q) =∫
MχJ−(q) =
∫MχI−(q)dµ
Since the relation << is open, it follows that χI−(p j ) converges pointwise toχI−(p) and the result then follows by Lebesgue’s dominated convergent theo-rem.
If now p << p j , we have that χJ−(p j ) converges pointwise to χJ−(p) as a conse-quence of the fact that the relation < is closed on globally hyperbolic manifoldand again f−(p j ) → f−(p) as j →+∞.
For a general sequence of p j , let ε > 0 and take points q1 and q2 such thatq1 << p << q2 and
f−(p) ≤ f−(q2) ≤ f−(p)+ε,
f−(p)−ε≤ f−(q1) ≤ f−(p).
for j large enough, q1 << p j << q2, so that
f−(p)−ε≤ f−(q1) ≤ f−(p j ) ≤ f−(q2) ≤ f−(p)+ε.
This proves the continuity of f−.
Step 2: f± are stricly monotone along causal curves.
Let p < q . Then q ∉ J−(p). (This follows from the strong causality condition).Since ≤ is closed on globally hyperbolic spacetimes, J−(p) is closed and thusthere exists an open neighborhood U of q which does not intersect J−(p). Itfollows that J−(q) \ J−(p) contains a non-empty open set and thus f−(q) >f−(p).
Step 3: For γ : [0, a) → M causal future inextendible, f+ γ(t ) → 0 as t → a.
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Let K ⊂ M be compact. We claim that for t large enough Kt = J+[γ(t )]∩ J−(K )is empty. Note that Kt is compact for every t , cf Lemma 10.14. Let C = K0 ∪γ(0). Then C is a compact set and γ start in C . From Lemma 10.9, there is at0 such that for t > t0, γ(t ) ∉C . Since γ(t ) ∈ J+[γ(0)] for all t ≥ 0, we concludethat γ(t ) ∉ J−(K ) for t > t0. Thus Kt is empty for t > t0. For any i , let now Ci bethe compact set defined by
Ci =i⋃
j=1suppφ j .
Since µ(M −Ci ) ≤ 2−i , we conclude that for any i , we have f+(γ(t )) ≤ 2−i , for tlarge enough.
Step 4: The function τ : M →R defined as
τ(p) = ln f−(p)− ln f+(p)
then satisfies all the properties of the theorem.
Definition 10.15. A continuous function which is strictly increasing along future di-rected causal curves is called a time function.
The time function we have just constructed still lacks regularity to be used inapplications. It is only continuous. Moreover, eventhough it is strictly increasingalong future directed causal curves, its gradient may not be timelike. We would liketo construct a smooth function τ such that Dτ is timelike and past-directed, whatwe will call a temporal function.
Definition 10.16. A temporal function τ is smooth function such that Dτ is timelikeand past-directed.
Exercise 10.7. • Prove that temporal function are time functions.
• Give an example of a smooth time function which is not temporal.
Our approach will be smooth out Geroch’s time function. For this, we will usecertain technical constructions presented in the next section.
10.8.2 Local constructions
First, we need the following lemma, whose classical proof is ommitted.
Lemma 10.18. Let V ⊂U ⊂ Rn be open sets. Let f ∈ C∞(U ) be such that f(x) > 0 forx ∈ V and f(x)=0 for x ∈ ∂V ∩U . Let g (x) = exp[−1/ f (x)], for x ∈ V and 0 elsewhere.Then g is smooth.
Let now p ∈ M and consider a normal neighborhood U of p and U = exp−1p (U ).
Recall the functions q : v → g (v, v) and q = q exp−1p . Let T +
p be the future timeconeat p and let V = expp T +
p ∩U . Note that q vanishes on ∂V and q < 0 on V and thuswe can define a function
fp (x) = exp[1/q(x)]
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for x ∈ V and fp (x) = 0 elsewhere. fp is smooth by the above lemma. Moreover, onV ,
D fp = grad fp =− fp
q2 grad q =−2fp
q2 P,
where P is the position vector field. Thus, D fp is a past directed timelike vector fieldon V and is zero everywhere else.
Lemma 10.19. Let S be an acausal Cauchy hypersurface. Let p ∈ S and let U be aconvex neighborhood of p. Then, there is a smooth non-negative function hp suchthat
1. hp (p) = 1,
2. hp has compact support contained in U ,
3. if r ∈ J−(S) and hp (r ) 6= 0, then gradhp (r ) is past directed timelike.
Proof. Let γ : (−ε,ε) → M be a future directed timelike curve with γ(0) = p. We claimthat for t < 0 close enough to 0, Kt = J+(γ(t ))∪ J−(S) is contained in U . First, the Kt
are compact by Lemma 10.17. Moreover, for s < t , Kt ⊂ Ks , i.e. the Kt are decreasingin t . Assume that for all t < 0, Kt is not included in U . Then, there exits an increasingsequence 0 > s j → 0 such that for all j , Ks j ∩U c 6= ;. Since p j ∈ Ks1 for all j , we canpass to a subsequence converging to r , and since U is open, U c is closed, so thatr ∈ Ks j ∩U c . Since γ(s j ) ≤ p j and ≤ is closed on globally hyperbolic manifold, wehave p ≤ r and even p < r (since r ∈U c and p ∈U ). Since r ∈ J−(S) and p ∈ S, thiscontradicts the acausality of S. We conclude that for t < 0 small enough, Kt ⊂U . Lett0 < 0 such that Kt0 ⊂U and let r0 = γ(t0). Since U is convex, we can construct thefunction fr0 as introduced above, which has a past directed gradient whenever it isnon-zero. Let K ⊂U be compact and containing Kt0 in its interior and letφ ∈C∞
c (U )be a cutoff function associated to K , i.e. φ= 1 on K and φ(U ) ⊂ [0,1]. Let H =φ f . Itis a smooth function with compact support in U and can therefore be extended by0 outside of U to a smooth function on M . Note that by definition of the functionfp (see the definition of the set V above), if r ∉ J+(γ(t0)), then f (r ) = 0. Thus, ifr ∈ J−(S) and H(r ) 6= 0, then H = f on a neighborhood of r , so that grad H is pastpointing and timelike. Finally, note that H(p) > 0, since p >> r0 and that φ(p) = 1.Thus, hp = H/H(p) has all the required property.
Proposition 10.8. Let S be an causal Cauchy hypersurface. Given an open neighbor-hood W of S, there exists a smooth function hW : M → [0,+∞) such that
1. The support of hW is included in W ,
2. hW |S > 12 ,
3. For r ∈ J−(S) and hW (r ) 6= 0, then gradhW (r ) is past pointing timelike.
Remark 10.7. Compared with the previous lemma, we are constructing a functionhW whose support and properties lies in a neighborhood of S instead of a neighbor-hood of some p ∈ S. Thus, we are gradually moving from local to more global proper-ties.
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Proof. Let d be the distance associated to a complete Riemannian metric on M .Then, Bρ(p), the closed ball of radius ρ centered at p, is compact due to the Hopf-Rinow theorem. Let p ∈ M , and define for l ∈N,
Kl = B l (p)−B l−1(p), Rl = Kl ∩S.
For each r ∈ S, fix a convex set Ur of diameter strictly less than 1 for d and containedin W . Let hr be the function constructed in the previous lemma for (U , p) = (Ur ,r ).Let Vr = h−1
r [(1/2,+∞)]. Since the hr are continuous with hr (r ) = 1, the Vr forms aopen cover of S. Since Rl is compact and contained in S, for any l , there exists a finitenumber of points, denoted rl ,1, ..,rl ,kl
∈ Rl such that the corresponding Vrl ,1 , ..,Vrl ,klforms an open cover of Rl . Note that if |l −m| ≥ 3, then Url ,i ∩Urm,i = ;, in view ofthe diameter constraint and the fact that d(rl ,i ,rm, j ) ≥ 2. Thus, the sum
h =+∞∑l=1
kl∑i=1
hrl ,i
defines a smooth function since each point has a neighborhood in which all but afinite number of the terms vanish. If x ∈ S, then h(x) > 1/2 since x must lies in someVrl ,i . Further, the support of h is the union of the supports of hrl ,i (i.e. we do not needto take a further closure), since the covering Url ,i is locally finite. Thus, the supportof h is included into W . If x ∈ J−(S) and h(x) 6= 0, then gradhrl ,i (x) is timelike andpast directed for every l , i such that hrl ,i (x) 6= 0 and vanishes for every l , i such thathrl ,i (x) = 0 (since hrl ,i ≥ 0, so that in that case x must be a minimum), so gradh(x) istimelike and past directed.
10.8.3 Smooth temporal function
Let τ be the continuous time function guaranteed to exist by Geroch’s theorem. Wewill denote its level set by St := τ−1(t ).
The idea will be to use the previous constructions to construct a replacement forτ near some St , and then glue the replacements to obtain a global temporal func-tion.
Definition 10.17. Let t− < ta < t < tb < t+ and S± = St± . A function σ : M → R iscalled a temporal step function around t (for (t−, ta , t , tb , t+) ) if it satifies
1. gradσ is timelike and past pointing in V = p ∈ M : gradσ(p) 6= 0,
2. σ(M) ⊂ [−1,1],
3. σ=±1 on J±(S±),
4. St ′ ⊂V for all t ′ ∈ (ta , tb).
Lemma 10.20. Let t− < t < t+. Then, there exists an open set U , such that
J−(St ) ⊂U ⊂ I−(St+)
and a function h+ : M → [0,+∞) whose support is contained in I+(St−) and such that
• If p ∈U and h+(p) > 0, then gradh+(p) is timelike and past pointing,
• h+(p) > 1/2 for p ∈ J+(St )∩U .
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Proof. Take h+ be the function h constructed in the previous Proposition with S = St
and W = I−(St+ )∩ I+(St−). For x ∈ St , h(x) > 1/2 and gradh(x) is past pointed andtimelike. By continuity, there exists an open neighborhood of x, Vx , such that thesame conditions on h and gradh holds in Vx . Then define U as the union of I−(St )and the sets Vx for all x ∈ St . (that J−(St ) ⊂U is a consequence of the fact that St is aCauchy hypersurface).
Lemma 10.21. Let t < t+ and U ⊂ I−(St+ ) be an open neighborhood of J−(St ). Then,there exists a smooth function h− : M → [−1,0] such that
• the support of h− is included in U ,
• if gradh−(p) 6= 0, then it is timelike, past directed.
• h−(p) =−1 for p ∈ J−(St ).
Proof. Reverse time orientation and construct a function h as in Proposition 10.8.Then we get a smooth function h : M → [0,+∞) whose support is included in Uand such that if h(p) > 0 for p ∈ J+(St ), then gradh(p) is timelike, future pointing.Moreover, h(p) > 1/2 for p ∈ St . Let h1 = −h and let φ : R→ [−1,0] be a smoothcutoff function such thatφ=−1 on (−∞,−1/2],φ′ > 0 on (−1/2,0),φ= 0, on [0,+∞).Defines h− = φ h1 on J+(St ) and h− = −1 on J−(St ), then h− has all the desiredproperty.
Proposition 10.9. Let t− < t < t+. Then, there exists a smooth functionσ : M →R sat-isfying the first three properties of Definition 10.17 and such that St ⊂ p ∈ M ,grad(σ) 6=0.
Proof. Let h+ and U be as in Lemma 10.20 and given this U , let h− be as in Lemma10.21. Then, h+−h− > 1/2 on U (since for p ∈ U , either p ∈ J+(St ) or p ∈ J−(St )).Thus, we can define σ as
σ= 2h+
h+−h− −1
on U and σ= 1 on the complement of U . We have
gradσ= 2h+gradh−−h−gradh+
(h+−h−)2
which at any given point is past pointing timelike or zero.
Corollary 10.5. Let t ∈ R and t± = t ± 1. Let t− < ta < t < tb < t+ and let K ⊂τ−1([ta , tb]) be compact. Then, there is a smooth function σ satisfying the first threeproperty of Definition 10.17 and such that K ⊂ p ∈ M ,grad(σ) 6= 0.
Proof. For each s ∈ [ta , tb], let σs be the function constructed in Proposition 10.9with t−, t , t+ replaced by t −1, s, t +1 and let Vs be the set on which gradσs is non-zero (i.e. the interior of the support of gradσ ). Then, the set Vs constitute an opencovering of K . Let s1, .., sk be such that the Vsi is a finite open covering of K anddefine
σ= 1
k
k∑i=1
σsi .
Then, σ has the desired properties.
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In the proof of the next theorem we will need the following lemma, whose proofis left as an exercice. Note that the reason why this is not a triviality is that a limit offuture-directed timelike vectors can be 0 or a null vector.
Lemma 10.22. Let vi be a sequence of future directed timelike vectors such that∑
vi
is convergent. Then,∑∞
i=1 vi is timelike.
Theorem 10.3 (Existence of temporal step function). Let t ∈ R and let t± = t ± 1.Given t− < ta < t < tb < t+, there is a temporal step function around t as in definition10.17.
Proof. Let G j be an increasing sequence of open sets of compact closure and suchthat M =∪ j G j . Let
K j =G j ∩ J+(Sta )∩ J−Stb .
K j is compact and can be used to get a function σ j as in Corollary 10.5. We willtake a sum of the σ j , but in order for the sum to converge we will need to divideeach σ j by certain weights. To define those, let
(Vi ,i ξ
)1≤i be a open covering by
coordinate charts. (Note that i in iξ labels the coordinate chart, not the individualcoordinate functions). We may assume that the Vi are locally finite and that thereexists another open cover by open sets Ui , such that Ui ⊂Vi for all i and the Ui havecompact closure. (Exercice: construct such open covers).
For any j , let A j > 1 be such that for all 1 ≤ i ≤ j , 0 ≤ m ≤ j and for all multi-indices |β| ≤ m,
supU i
∣∣∣∣∣ ∂βσ j
∂m iξ
∣∣∣∣∣≤ A j ,
i.e. the partial derivatives of order m of σ j are uniformly bounded by A j on U i forthe j first coordinate systems.
Let now σ be defined by
σ=+∞∑j=1
1
2 j A jσ j .
Let p ∈ M . Then p ∈ Ui . To prove that σ is C l , split the series into j > i , l andj ≤ i , l . The j ≤ i , l are finite, so poses no threat to regularity.
For j > l , i ,σ j
2 j A jand all its derivatives of order ≤ l with respect to the coordinates
iξ are bounded by 12 j , so we have uniform converges. It follows that σ is smooth.
We still need to normalize σ to satisfy all properties of 10.17. For this, note thatσ = σ− for some constant σ− < 0 on J−(S−) (since all the σ j = −1) and similarlyσ = σ+ > 0 on J+(S+). Let ψ : R→ [−1,1] be such that, ψ is smooth, ψ(t ) = −1 fort ≤σ−, ψ(t ) = 1 for t ≥σ+ and ψ′ > 0 for t ∈ (σ−,σ+). Then ψσ has all the desiredproperties.
Theorem 10.4. There exists a smooth time function T : M → R, whose gradient istimelike and past pointing and T tends to ±+∞ along any future/past inextensiblecausal curve. In particular, the level sets of T are smooth spacelike Cauchy hypersur-faces.
Proof. Let tk = k/2, tk,± = tk ± 1, tk,a = tk ±−1/2, tk,b = tk ±+1/2. Let σk be thefunction just constructed, with t = tk , t± = tk,±, ta = tk,a , tb = tk,b . Note that each
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p ∈ M is such that τ(p) ∈ (tk,a , tk,b) for some k. Define
T =σ0 ++∞∑k=1
(σ−k +σk ).
Note that if k ≥ 3, then σ−k (p) +σk (p) = 0 if −k/2 + 1 ≤ τ(p) ≤ k/2 − 1 (becauseσ−k (p) = 1, σk (p) = −1 on this interval). Thus, for any p ∈ M , there is a neighbor-hood of p such that only a finite number of the terms in the above sum are non-zeroon that neighborhood. It follows from the definition and the fact that each p is suchthat τ(p) ∈ (tk,a , tk,b) for some k that the gradient of T is timelike and past directed.Let now γ : (t−, t+) → M be a future directed inextensible causal curve. From the gra-dient properties of T , T γ is strictly increasing along γ. Let m ≥ 1. Since γ inter-sects each Cauchy hypersurfaces St , there exists sm ∈ (t−, t+) such that τ(γ(sm)) = m.Let l = 2(m +1). We have (σ−k +σk )[γ(sm)] = 0 for k ≥ l . Thus,
T (γ(sm)) =[σ0 +
l∑k=1
(σ−k +σk )
](γ(sm).
Since m ≥ 1, σ0[γ(sm)] = 1. Futhermore, (σ−k +σk )(γ(sm) ≥ 0 for all k ≥ 1, sinceσ−k (γ(sm)) = 1 and σk (γ(sm)) ≥−1 . Finally, for 1 ≤ k ≤ 2(m −1), we have
(σ−k +σk )(γ(sm) = 2
and thusT (γ(sm)) ≥ 4(m −1)+1.
Combined with the monotonicity property of T (γ), we have obtained T (γ(s)) →+∞, as s → t+.
The above theorem says in particular that given a globally hyperbolic space-time, there exists a foliation by smooth spacelike Cauchy hypersurface. On the otherhand, we know that a spacetime admitting a Cauchy hypersurface, and thus in par-ticular, a smooth spacelike Cauchy hypersurface, is globally hyperbolic. The nexttheorem, whose proof is omitted, then states the existence of a foliation by smoothspacelike Cauchy hypersurface such that one of the leaf coincides with the originalCauchy hypersurface.
Theorem 10.5. Let S be a smooth spacelike Cauchy hypersurface. Then, there exists atemporal function τ such that τ−1(0) = S and τ goes to ±∞ along future/past directedinextendible causal curves.
11 The domain of dependence in globally hyperbolic space-time
The aim of this section is to prove the following statement.
Theorem 11.1 (Domain of dependence). Let (M , g ) be a globally hyperbolic space-time and let t be a temporal function as in Theorem 10.4. Let S = t−1(0) and letΩ⊂ Sbe open. Let u be a solution to äg u = 0 such that u and Du vanishes on Ω. Thenu = 0 on D+(Ω). In particular, taking Ω = S, then we have proven uniqueness of thesolution to the linear wave equation.
Before proving the theorem, we will need a few more geometric statements, aswell as a local version of it.
106
11.1 Three extra geometric lemmas
Lemma 11.1. Let (M , g ) be a smooth globally hyperbolic spacetime with a smoothspacelike Cauchy surface S. Then, for every p ∈ S, there is a neighborhood Up of psuch that for every q ∈Up ∩ J+(S), there is a normal neighborhood Vq of q such thatJ−(q)∩ J+(S) is compact and contained in Vq .
We leave its proof as an exercice, but note that one cannot just take any convexneighborhood of p, for the condition J−(q)∩ J+(S) ⊂Vq is not always satisfied. (Fora full proof, consult [Rin09], p135 ).
We will also need (again for proofs, see [Rin09, Chapter 12].)
Lemma 11.2 (The intersection of transverse manifolds is a manifold). Let N1 andN2 be two n dimensional spacelike submanifolds. Asssume for each p ∈ N1 ∩N2, thenormals to N1 and N2 are linearly independent. Then, N1∩N2 is an n−1 dimensionalspacelike submanifold.
Lemma 11.3. Let N1 and N2 be two n dimensional spacelike submanifolds of (U ⊂Rn+1, g ), for some Lorentzian metric g defined on U . Assume N1 and N2 are such thattheir normals are linearly independent on N1 ∩N2, so that from the previous lemma,P = N1∩N2 is a n−1 dimensional manifold. Assume moreover that P is compact. Fixn1 and n2 be the normals to N1 and N2. Then, their restrictions to P give two vectorfields on P such that at every p ∈ P, n1(p),n2(p) ∈ Tp M⊥. Define
f : P ⊗R2 →Rn+1
byf (p, t , s) = p + tn1(p)+ sn2(p).
Then f is smooth and there is an ε > 0 such that f restricted to P ×Bε(0) is a diffeo-morphism onto a neighborhood of P.
11.2 A local version of the domain of dependence property
Lemma 11.4. Let S be a Cauchy surface and p ∈ J+(S) \ S. Assume that there is anormal neighborhood of p Vp such that J−(p)∩ J+(S) is compact and contained inVp . Then, if u is a smooth solution to the wave equation
äg u = 0
and u an Du vanishes on J−(p)∩S, u = 0 on J−(p)∩ J+(S).
Proof. Let Vp = exp−1p (Vp ). We consider at p, the hyperquadric q−1(c) ∈ Tp M , for c <
0. It has two connected components, and we consider the one corresponding to thepast directed timelike vectors, denoted Qc . Let Qc be the image by the exponentialmap of Qc ∩ Vp . By construction Qc ⊂ I−(p). Similarly, we denote by Q0 the pastdirected local null cone at p. Let D = J−(p)∩ J+(S) and Dc = J−(Qc )∩ J+(S). Weleave as an exercice to verify that the following statement holds (Draw a picture ! )
1. Dc =⋃γ≤c Qγ∩ J+(S),
2. The interior of Dc is I−(Qc )∩ I+(S) and its boundary is Qc ∩ J+(S)∪ J−(Qc )∩S.
3. If cl → 0, cl < 0 for all l , then int D ⊂⋃l Dcl ⊂ D .
107
Let ρ be a Riemannian metric on V and let d be the associated topological met-ric. Let ε> 0 and define
Rε = r ∈ S ∩D : d(r,Q0) < ε.
Then, Rε is an open neighborhood of Q0 ∩S in S ∩D . Let Lε = S ∩ J−(p) \ Rε. SinceD is compact and S is closed, we have D ∩ S compact and hence, Lε is compact.Moreover, for every r ∈ Lε, exp−1
p (r ) is timelike (because r does not belong to some
Q0 but lies in J−(p) ). Consequently, exp−1p (Lε) is a compact subset of the interior of
the past line cone in Tp M . Hence, for c < 0 small enough, Lε does not intersect Qc
and Qc and S must intersect in Rε. Let T be a smooth unit normal to S. Then, T istimelike. Another vector field in Rε is provided by the position vector field P , whichwe recall is normal to any of the Qc . For ε> 0 small enough, we claim that P and Tare linearly independent in Rε. Since P and T are non-zero vector fields on Rε andRε is compact, ρ(T,T ) and ρ(P,P ) are uniformly bounded above and below on Rε.On the other hand, g (T,T ) =−1 and g (P,P ) tends to zero as ε→ 0. This implies thelinear independence.
Consequently, for c < 0, every point in Qc ∩S is such that the normal to Qc andS at that point are linearly independent. Since Qc and S are smooth spacelike ndimensional manifolds, Qc∩S is a smooth n−1 dimensional submanifold. Moreoverit is compact (exercice).
Let u be the solution assumed to exist in the statement of the lemma. Let T beits energy-momentum tensor, i.e.
Tαβ = DαuDβu − 1
2gαβ(g (Du,Du)).
Let f =−1/2q and N =−P , where P is the position vector field. Note that grad f = N .Let J := JN
β= TαβNβ be the current associated to N and let
η=−ek f |u|2N
andJ = ek f J
be a weighted version of it. (The weights allows to absorb some error terms, other-wise, we would need an extra Gronwall argument to conclude). The constant k wilbe chosen later.
Note that in Dc , N is a future directed timelike vector field and that on Qc∩I+(S),it is the outward pointing normal relative to Dc . Recall that
DαTαβ =äg u.Dβu
anddiv J =äg N (u)+Tαβ πN
αβ.
Let e = 12 |u|2 +
∑α |∂αu|2.
Then,|div J | ≤Ce,
for some C > 0.We also have
div η=−2ek f uN (u)−ek f |u|2di v N −kek f |u|2g (N , N ).
108
Since N is timelike on Dc , there exists a constant c0 > 0 such that
div η≥ ek f (kc0|u|2 −Ce
).
Moreover,div J = ek f (div J +kQ(N , N ))
and again, since N is timelike on Dc (uniformly), there exists a constant c1 suchthat
c0|u|2 +T (N , N ) ≥ c1e.
Hence,div η+div J ≥ ek f (kc1 −C )e.
For k large enough, the left hand side is therefore positive and controls ek f e. More-over,
g (N , J )
g (N , N )= ek f T (N , N )
g (N , N ),
g (N ,η)
g (N , N )= e−k f |u|2.
Both of these quantities are non-positive. Recall that by Stokes theorem, if (M , g ) isLorentzian manifold with a smooth spacelike boundary ∂M and if N is the outwardunit-normal, then for ξ is smooth vector field with compact support∫
Mdiv ξ=
∫∂M
g (ξ, N )
g (N , N )(27)
where both integrals are taken with respect to the natural (induced) volume formson M and ∂M . If we could apply this directly in our setting we would be done, sinceboth sides of the equation would have different signs.
The trouble is that our domain Dc has corners at the intersection of Qc with S,so we are not just in the setting to apply the above.
We will use Lemma 11.3 above.Using the lemma, we get a smooth map
h : Qc ∩S ×Bε(0) →V
which is a diffeomorphism onto its image and contains an open neighbourhood ofQc ∩S, asssuming that ε> 0 is small enough. Let χ ∈C∞
c (R2) be such that χ(x) = 1 for|x| ≤ 1/2 and χ(x) = 0 for |x| ≥ 3/4 and let χ(δ/x) = χδ(x). For δ≤ ε, we can considera function
φδ : Qc ∩S ×Bε(0) → R,
(p, x) → ψ(p, x) :=χδ(x)
Then,ψ=φh−1 is a smooth function defined on some neighbordhood of Qc∩S andwe can extend it by 0 outside to get a smooth function on V . Moreover, the volumeof the support of ψδ can be estimated by Cδ2 for some constant C and
|∂ψδ| ≤Cδ−1.
Now D ′c = Dc \ S ∩Qc is a smooth manifold with boundary and (1−ψδ)X has
compact support on this manifold. Applying the above divergence formula (27), weget ∫
D ′c
div((1−ψδ)X
)=−∫
D ′c
Xα∂αψδ+∫
D ′c
div X −∫
D ′c
ψδdiv X .
109
The first and the last term converges to 0, so that the middle term converges tothe divergence of X on Dc . The boundary integrals similarly converges to whatthey should. It follows that the above divergence formula (27) is valid even in thepresence of "corners" (not that it was important that they were of co-dimension 2!)which concludes the proof.
11.3 Proof of the domain of dependence
We now move with the proof of Theorem 11.1.
Proof. Note first that D+(Ω) ⊂ int (D+(Ω)). Let p ∈ int (D+(Ω)). We have K = J−(p)∩D+(Ω) compact by Lemma 10.17. Recall that t is a smooth temporal function andthat S = t−1(0). For any interval I and t0 ∈R, we define the sets
RI = t−1(I )∩K , Rt0 = t−1(t0)∩K , St0 = t−1(t0).
Note that if I is compact then so is RI and that Rt0 and St0 are compact. Let Iε =[t0 −ε, t0 +ε].
Note that, for any t0 such that Rt0 is not empty and any open set U containingRt0 , by compactness of Rt0 , there exists ε> 0 such that RIε ⊂U .
Let now 0 ≤ t0 ≤ T := t (p). For any q ∈ Rt0 , there is an open neighborhood Uq asin the Lemma 11.1, using St0 as a Cauchy surface. The Uq s form an open cover ofRt0 and by compactness of Rt0 , there is a finite subcovering Uq1 , ..,Uql . Let U be theirunion and let ε be small enough so that RIε ⊂U . Assume that u and Du vanishes onRt0 . Let q ∈ Rs for some s ∈ [t0, t0 + ε]. There there is a normal neighborhood of qsuch that the conditions of Lemma 11.4 holds, which implies that u and Du vanisheson Rs . Thus, the set of s ∈ [0,T ) such that u and Du equal zero on R[0,s] is non-empty,open and closed in [0,T ), which proves the theorem.
12 Symmetric hyperbolic systems
12.1 Solving PDEs: representation formula vs abstract methods
There are many ways to solve linear wave equations. Here, we will go through a stan-dard method which relies essentially on energy estimates and a duality argument.The advantage of this method is that it is an abstract method, that works in manydifferent situations and that it sort of highlights the importance of a priori estimates.Another standard method is via the construction of paramatrices or representationof solutions. For instance, for the free wave equation in Minkowski space,
äφ := (−∂2t +∆
)φ0,
one can use the Fourier transform to get an explicit representation ofφ given appro-priate initial data. In a curved space, one cannot use the Fourier transform, but onecan still obtain explicit various representation of the solutions37 under certain con-ditions. See for instance [Fri75] in the case of spacetimes for which the exponentialmap is always a global diffeomorphism. The advantage is that from the representa-tion formula one typically get more information than just existence. The drawbackis that the more you ask about the solution, the less likely it is to apply in a generalsituation.
37Often, the representation is only an approximate solution so that one typically have errors terms. Onecan then get rid of the error terms by iterating in a second step.
110
12.2 Symmetric hyperbolic systems on Rn+1
In this section, we consider an equation of the form
L(u) := Aµ∂µu +Bu = f , (28)
u(0, .) = u0, (29)
where u is anRN valued function on defined onRn+1 (or a subset), the initial data u0
is taken smooth for simplicity, f is smooth and decay fast at infinity in x (say f (t , .) isSchwarz for all t ) and the Aµ are smooth N ×N matrix valued functions defined onRn+1 such that Aµ are symmetric for each µ and A0 is positive definite with a uniformlower bound, i.e. ∃c > 0, such that
A0(ξ,ξ) > c|ξ|2, ∀ξ ∈RN .
The operator L is then said to be symmetric hyperbolic.Recall that H k = H k (Rn) is the space of tempered distributions u on Rn such
that its Fourier transform u is a measurable function (we identify the function andits distribution) and such that
||u||H k := ||(1+|ξ|2)k/2|u|||L2 <C .
H k with the norm ||.||H k is then a Hilbert space. Recall also that for k ≥ 0, we canidentify H k with the space of L2 functions which have k weak derivatives in L2.
Recall also that (H k )∗ can be identified with H−k in the standard way.We will need the following lemma, whose proof is left as an exercice (hint: use
duality).
Lemma 12.1. If u if Schwarz and φ is C∞ with uniform bounds on φ and its deriva-tives, then
||φu||H k . ||u||H k ,
even for negative k.
12.3 A priori estimates for symmetric hyperbolic systems
Lemma 12.2 (Estimates for k ≥ 0). Let u be a solution of (28) which is smooth on[0,T ]×Rn and such that, for k ≥ 0, u(t , .) ∈ H k , for all t ∈ [0,T ]. Let Ek be defined by
Ek [u] = 1
2
∑|α|≤k
∫Rn
A0(∂αu,∂αu)d x.
Then, we have the following estimate
E 1/2k (t ). E 1/2
k (0)+∫ t
0|| f (s, .)||H k d s.
Proof. We do the proof for k = 0, the other follows easily after commuting the equa-tion by ∂α and noting that the resulting equation can be put under the same form.
We have
d
d t1/2
∫x
A0(u,u)d x = 1/2∫
x∂t A0(u,u)d x +
∫x
A0(∂t u,u)d x,
= 1/2∫
x∂t A0(u,u)d x −
∫x
ut(
Ai∂i u +Bu − f)
d x,
= 1/2∫
x∂t A0(u,u)d x +1/2
∫x
(ut∂i (Ai )u − (Bu,u)− ( f ,u)
)d x,
111
so that the lemma follows from the lower bounds on A0 and the upper bounds on Aand f and B .
Since we have uniform lower bounds on A0 and uniform H k bounds on f byassumptions, we obtain immediately
Corollary 12.1. Under the same assumptions, we have uniform ||.||H k bounds on ufor k ≥ 0.
Because of the duality arguments we will use later, we also need uniform esti-mates on u in H k for k < 0.
Lemma 12.3 (Estimates for k < 0). Let u be a solution of (28) which is smooth on[0,T ]×Rn , satisfying uniform Schwarz bounds, for all (t , x) ∈ [0,T ]×Rn ,
(1+|x|i )|∂βu|(t , x) ≤Ci ,β.
Let k < 0 be a negative integer. Then, we have the estimate
||u(t , .)||H k . ||u(0, .)||H k +∫ t
0|| f (s, .)||H k d s.
Proof. Let U (t , .) be defined by
U (t , .) = (1−∆)k u(t , .),
where (1−∆)k for k negative is to be understood as the Fourier multiplier (1+|ξ|2)k .It follows from the assumptions on u that both U and LU satisfy Schwarz bounds,so that we can apply the results of the previous corrolary. Thus
||u(t , .)||H k = ||U (t , .)||H−k . ||U (0, .)||H−k +∫ t
0||LU (s, .)||H−k d s
. ||u(0, .)||H k +∫ t
0||LU (s, .)||H−k d s.
It remains to estimate LU in H−k . For this, we observe that
f = Lu = (1−∆)−k LU + [L, (1−∆)−k ]U
yielding||LU ||H−k ≤ || f ||H k +||[L, (1−∆)−k ]U ||H k .
Note that[L, (1−∆)−k ] = ∑
|α|≤2|k|aα∂
α
where α are multi-indices such that ∂α contains at most one t derivative and aα aresmooth bounded coefficients depending only on A and B .
The previous analysis already brings, in particular, a global uniqueness state-ment: if u and v agree at t = 0, they agree for all times (just apply Lemma 12.2 withk = 0). However, one can prove a stronger result, a local uniqueness statement.
12.4 A rough local uniqueness statement
Lemma 12.4 (local uniqueness: Domain of dependence). Let L be as above withAµ ∈C 1, A0 positive definite and bounded from below, B ∈C 0 and f = 0 on some slab[T1,T2]×Rn . Let u be a solution to (28) such that u(0, .) = 0 on some some ball Bx0 (R)with R > 0. Then, there exists a c > 0 depending only the bounds on A such that u = 0on I−(p)∩ [T1,T2]×Rn , where p = (R/c, x0) ∈ Rn+1 and I−(p) is the interior of a pastnull cone for the Minkowski metric with the speed of light c.
Proof. We multiply the equation by e−kt u for some k large enough to get
∂α[e−kt Aα(u,u)] = e−kt (−k A0(u,u)+∂αAα(u,u)−2B ||u||2)and integrate this over a domain of the form D = I−(p) ∩ [0,T2] ×Rn , where p =(R/c, x0) and c will be chosen large enough so that R/c < T2. Moreover, here I−(p)refers to the interior of the past null cone of p associated with the Minkowski metric
ηc =−c2d t ⊗d t +δi j d xi ⊗d x j .
We then use the usual Stokes’ theorem in Rn+1 to relate the bulk terms to theboundary terms. The integral over ∂D ∩ t = 0 vanishes in view of the initial datahypothesis. The other boundary is the truncated cone given by the equation
c
(t − R
c
)= |x −x0|, t > 0,
and the vector n = (nα) =(
c
− (x−x0)i
|x−x0|
)is an outgoing normal. (ex: rewrite this using
differential geometric language, then n is actually a one-form...)Thus this boundary term give a contribution proportional to nαAα(u,u) where
n0 can be made arbitrary large by choosing c large enough. Choosing c large enough,it follows that this boundary term is non-negative. Choosing k large, the bulk termis non-positive, which implies that all terms must vanish, i.e. u = 0.
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Remark 12.1. If we apply the above statement to the case of wave equation, this formof the domain of dependence is weaker than the one we proved in the previous section,because we are proving a domain of dependence using an auxiliary Minkowski metricrather than the true geometry of our spacetime. Essentially, we are saying that the truelightcone can always fit into a larger auxiliary flat lightcone, or yet in another words,we overestimated the speed of light.
12.5 Existence by duality
We are now in a position to prove existence of solutions.
Theorem 12.1 (Existence). Let ST = [0,T ] ×Rn and assume that L is above, withAµ and B ∈ C∞ with bounded derivative A0 positive definite with a uniform lowerbound. Assume for simplicity that u0 ∈C∞
c , i.e it has compact support. Then, there isa unique solution u on ST to (28). Moreover, for each t , u(t , .) = 0 outside from a fixcompact KT .
Proof. Let L be as above and define L? be the formal adjoint of L
L?u =−∂t (A0u)−∂ j (A j u)+B t u.
Then, L? is a symmetric hyperbolic operator, so in particular, we have, for any kinteger and for every test function φ ∈C∞
c ((−∞,T )×Rn),
||φ(t , .)||H−k ≤C∫ T
t||L?φ(s, .)||H−k d s, (30)
for t ∈ [0,T ], by applying the estimates of Lemma 12.3 with t replaced by T − t .Let X be the Banach space
X = L1([0,T ], H−k
).
Note that L?φ can be viewed as a member of X so we let M be the subspace of Xcomposed of elements of the form L?φ for φ as above, i.e.
M = L?(C∞
c
((−∞,T )×Rn))
.
Note that for any elementψ of M , there is a uniqueφ, so thatψ= L?φ, in view of theabove a priori estimate.
For any f ∈ L1([0,T ], H k
), we consider a functional F (:= F f ) on M , defined as
follows. For ψ= L?φ ∈ M ,
F (ψ) =<φ, f >:=∫ T
0(φ(t ), f (t ))L2 d t .
Then F is a linear functional on the subspace M and it follows from (30) that it is abounded functional on M in the norm of X . By the Hahn-Banach theorem, it admitsa continuous extension to the whole of X satisfying the same bound. By duality38,there exists a u ∈ L∞ (
[0,T ], H k)
such that, for any ψ ∈ X ,
F (ψ) =<ψ,u > .
38Recall that (L1)? is isomorphic to L∞, while (L∞)? is not isomorphic in general to L1.
114
In particular, for all ψ= L?φ ∈ M ,
F (ψ) =∫ T
0(φ(t ), f (t ))L2 d t =
∫ T
0(L?φ(t ),u(t ))L2 d t .
Thus u is a solution in the sense of distributions of the equation (since we haveassumed that u(0, .) = 0). From the equation, the weak time derivative ∂t u solves
∂t u =−(A0)−1 Ai∂i u −Bu − f ∈ H k−1.
Thus, we have ∂t u,u ∈ L∞ ((−∞,T ); H k−1
), which implies that39 u ∈C 0
((−∞,T ); H k−1
).
Repeating the argument, we can obtain that u ∈C 1((−∞,T ); H k−2
). In particular, if
f is smooth, this implies (using k arbitrary large) that u is smooth. To solve the equa-tion with non-vanishing data, assume first that u(0, x) ∈C∞
c and let u0(t , x) = u(0, x)so that u0 is a spacetime function with the right-Cauchy data. Then, consider themodified equation Lv = f −Lu0.
Corollary 12.2. Let u0 ∈ C∞ and f ∈ C∞. Assume that the operator L is as above,there exists a unique solution to (28)
Proof. Use the previous existence result and the local uniqueness together.
12.6 Applications to linear wave equations on Rn+1
We now consider a linear wave equation in Rn+1 with a curved Lorentzian metric gof components gµν, given by
gµν∂µ∂νu +bα∂αu + cu = f , (31)
u(t = 0, .) = u0, (32)
∂t u(t = 0, .) = u1, (33)
where bα,c,u0 and u1 are C∞ functions. We assume that g is uniformly Lorentzian,as in Definition 3.5.
Without loss of generality, assume that g 00 = −1 (otherwise, just divide). Letv = (∂αu,u) be a Rn+2 vector field over Rn+1 and define the symmetric n +2×n +2matrices by (A0)i j = g i j , (A0)n+1,n+1 = (A0)n+2,n+2 = 1, (Ak )i ,n+1 = (Ak )n+1,i =−g i k ,(Ak )n+1,n+1 =−2g 0k and all other components vanishes. Let moreover d n+1,i =−bi ,d n+1,n+1 = −b0, d n+1,n+2 = −c, d n+2,n+1 = −1, hn+1 = f . Then, (31) is equivalent tothe equation on v
Aα∂αv +d v = h. (34)
Exercise 12.1. Show reciprocally that if v solves the above equation and the dataverifies ∂i vn+2 = vi , then u = vn+2 solves the original wave equation.
From our previous analysis, we therefore obtain
Theorem 12.2. There exists a unique C∞ solution u to (31) satisfying the initial con-dition.
As well as a domain of dependence property
Proposition 12.1. With the above notation, if f = 0 and if the initial data vanisheson some BR (x0), then, there exists a c > 0, such that u = 0 on I−(p)∪ t ≥ 0, wherethe set I−(p) is constructed from the Minkowski metric with speed of light c and p =(R/c, x0).
39Recall that we denote by u both a distribution and a function representing u. Thus ui sC 0, meansthere exists a C 0 function whose distribution defined by it equal u.
115
13 Existence theory for the wave equation on globallyhyperbolic spacetimes
In Section 11, we proved the domain of dependence property for solutions of thewave equation on globally hyperbolic spacetimes but we still have not proven exis-tence of actual solutions. In the previous section, we prove an existence result whichholds for uniformly Lorentzian metrics.
We will use the following lemma to reduce the general case to the case of uni-formly Lorentzian metrics. For the proof, see [Rin09], p141.
Lemma 13.1. Let (M , g ) be a globally hyperbolic spacetime and t a smooth temporalfunction whose level sets St are Cauchy hypersurfaces. If p ∈ St , then there exists ε> 0and open neighborhoods U and W of p such that
1. W ⊂U and is compact.
2. if q ∈W and τ ∈ [t0 −ε, t0 +ε], then the compact set J+(Sτ)∩ J−(q) is containedin U .
3. There is a coordinate system on U , φ = (xα) such that x0 = t and the metriccomponents gαβ in this coordinate systems are uniformly Lorentzian on U .
4. For any compact K ⊂ U , there exists a uniformly Lorentzian metric h globallydefined on Rn+1 such that h coindices with g φ−1 on φ(K ).
We then have
Theorem 13.1. Let (M , g ) be a globally hyperbolic spacetime and t a temporal func-tion as above. Then, given smooth functions φ0, φ1 defined on S := St0 , there exists asmooth solution φ of
äφ = F,
φSt0= φ0, N (φ)St0
=φ1,
where N is one of the unit normals of St0 .
Proof. Let us assume that N is the future unit normal and construct the solution tothe future of S. First, we assume that φ0, φ1 are compactly supported in K1 ⊂ S andthat F is compactly supported in K2 ⊂ M . Let t1 > 0 and define Rt1 to be the closedset of all qs such that 0 ≤ t (q) ≤ t1, i.e. Rt1 = t−1 ([0, t1]). Let K3 = K2 ∩Rt1 . Note thatK3 is compact. The union of all I+(p) for p ∈ I−(S) is an open covering40 of K1∪K3, soit admits a finite subcovering consisting of I+(pl ), 1 ≤ l ≤ k < +∞, pl ∈ I−(S). Notethat K4 =⋃k
l=1 J+(pl )∩ J−(St1 ) is compact41. By the domain of dependence theorem11.1, any solution in Rt1 must vanish in Rt1 \ K4. Thus, we only need to define oursolution φ in the compact K4 ∩Rt1 . Let K4(t ) = K4 ∩St . Note that if K4(t ) ⊂U for Uopen then there exists an ε> 0 such that K4(s) ⊂U for s ∈ [t − s, t + s].
Let 0 ≤ τ ≤ t1 and assume that we either have a solution on Rτ or on Rs for all0 ≤ s < τ. For all p ∈ K4(τ), there are neighboorhoods Wp , Up and an εp > 0 as inLemma 13.1. By compactness, there is a finite number of points pi , 1 ≤ i ≤ l such
40Note that since K1 ⊂ S, the union of all I+(p) for p ∈ S does not cover S and hence does not cover K1.41It would be more intuitive to consider the Cauchy development of the set where both the data and
the source vanishes.
116
that the Wpi forms an open covering of K4(τ). Let 0 < ε≤ min1≤i≤l (εpi ) be such thatthe Wpi form an open cover of K4(s) for all s ∈ [τ− s,τ+ s] and let s1 ∈ [τ− ε,τ] suchthat there is a solution up to and including s1.
Let s ∈ [s1,τ+ ε] and p ∈ K4(s). Then Kp = J−(p)∩ J+(Ss1 ) is compact and, byProperty 2 of the lemma, contained in one of the charts, say (Upi ,φ). Consider thewave equation on Upi with initial data given on Ss1 . We can truncate the data so thatit coincides with induced data on Kp ∩Ss1 and vanishes outside from some open setincluded with Upi . Similarly, we can truncate all lower order coefficients and sourceterms in the wave equations, keeping only the principal symbol gαβ∂α∂β. By thethird property of the lemma, we can moreover, extend g outside of Kp to a globallydefined uniform Lorentzian metric on Rn+1. We then obtain a linear wave equationas in the previous section to which there exists a smooth global solution. This givesus a solution on Kp . For any q ∈ I−(p)∩J+(Ss1 ) =Vp , we definedφ to be this solution.
For r ∈ Vp ∩Vq , we have two candidate definitions for φ at r . Since J−(r ) ∩J+(St1 ) ⊂ Vp ∩Vq , it follows from our domain of dependence property that the twopossible definitions coincide, so that our solution is well defined.
Let nowO1 =
⋃p∈K4(s), s∈[s1,τ+ε]
Vp .
Note that the interior of O1 contains K4(s) for any s ∈ (s1,τ+ ε). Let O2 be the set qsuch that s1 ≤ t (q) < τ+ ε and q ∉ K4. Note that if q ∈ O2 and t (q) > s1 then q is inthe interior of O2 and similarly for O1. In O1, we have already define φ and in O2, wewould like to defined it to be 0. If q ∈ O1 ∩O2 and t (q) > s1, then φ and its normalderivatives vanish at J−(q)∩Ss1 and F vanishes in J−(q)∩ J+(Ss1 ). Moreover, thereis an r > 0 such that q ∈ Vr ⊂ O1. By uniqueness, the solution defined on O1 musttherefore vanish at q .
Thus, given a solution on Rτ or on Rs for any s < τ, we get a solution on Rτ+ε, forsome ε> 0. Thus, the set of s ∈ [0,+∞) such that there is a solution on Rs is closed,open and non-empty, i.e. equal [0,+∞).
Finally, we need to remove the compact support assumption. Let p ∈ J+(S).Then, Kp = J−(p)∩ J+(S) is compact. Truncate the data and F to be zero outsidesome compact set and such that the induced data coincides with the original dataon Kp . We then get a smooth solution φ′. Defined φ to be φ′ on I−(p)∩ J+(S) = Vp .Again, ifr ∈Vp ∩Vq , then by uniqueness, the two possible solutions must coincide atr .
14 Local existence for the vacuum Einstein equations
We have seen in Section 6.3 that, using wave coordinates, the Einstein equationsreduce to a system of quasilinear wave equations. Given proper initial data, thiskind of system can be solved locally in time, cf [Sog95b, Chapter I.4].
What remains to be done to turn this into a proper existence result for the Ein-stein equations is to explained how to set-up the initial data and how to propagatethe wave gauge.
First a definition
Definition 14.1. An initial data set for the vacuum Einstein equations is a triple(Σ,h,k) where (Σ,h) is a Riemannian manifold and k is a symmetric (0,2) tensor so
117
that the following constraint equations are satisfied
1
2S(h)−|k|2 + (trhk)2 = 0,
divk −d(trhk) = 0,
where S(h) denotes the scalar curvature of (Σ,h) and |k|2 = ki j k i j .
Comparing the above equations with (24)-(25), we see that we want to think of kas the 2nd fundamental form of Σ. However, at the moment, Σ is not a submanifoldof M , in fact, there is no manifold M yet.
Let us now define the concept of a solution, or developement associated to someinitial data.
Definition 14.2. Let (Σ,h,k) be an initial data set. Then, a developement of (Σ,h,k)is solution (M , g ) to the vacuum Einstein equations together with an embedding of Σas a smooth hypersurface of M,
ψ :Σ→ M ,
such that (h,k) coincide after pullback, with the first and second fundamental formof the embedding.
Let us do a little bit of functions counting. Assume Σ is of dimension n. Then,M is of dimension n + 1. We are seeking for a Lorentzian metric g , which we canthink as (n +1)(n +2)/2 functions in view of the symmetry of g . If we are thinkingof solving the Einstein equations using the reduced equations in wave coordinates,we need data for g and ∂t g , which is thus (n +1)(n +2) functions. These functionsare not independent because of the n +1 wave gauge conditions. So we have only(n +1)2 free functions. (h,k) provide n(n +1) functions, but we should also take inaccount the n +1 constraint equations, so that we really have
n(n +1)− (n +1) = (n −1)(n +1)
independent functions. Thus, we are missing 2(n+1) initial data information. Theseare fixed by the choice of wave coordinates: in order to construct a function satisfy-ing the wave equation, one needs data for it and its derivatives and there are (n +1)coordinate functions.
Consider now an initial data set (Σ,h,k). Let(U , (xi )
)be a local coordinate sys-
tem of Σ. We will solve the Einstein equations on an open subset M of I ×U , whereI is a an open interval containing 0. Moreover, M will contain the submanifoldP = t = 0×U . Note that such an M can be given a system of global coordinates(t , xi ).
The metric g will be obtained by solving the reduced Einstein equations. Sincethis is a system of quasilinear wave equations, we need initial data for g and ∂t g .The initial data will be imposed on the hypersurface P = t = 0×U . We define themin local coordinates as
gi j = hi j , g00 =−1, g0i = 0.
Note that with this choice, if we assume that we have already constructed a solution,the vector field ∂t is a unit normal to P . The second fundamental form is then given
118
by
ki j = g (∂xi ,D j∂t ) = giαΓαj t
= giα1
2gαβ
(g jβ,t + g tβ, j − g j t ,β
)= hi k
1
2hkl g j l ,t
= 1
2g j i ,t .
Thus, we define gi j ,t := 2ki j . We are still missing the data for g00,t and g0i ,t . Again,we imagine we already have a solution in wave coordinates. Then, the wave coordi-nate condition gαβΓ0
αβ= 0 give us on P ,
gαβΓ0αβ = −Γ0
00 +hi jΓ0i j
= −1
2g 0α (
gα0,t + g0α,t − g00,α)+hi j 1
2g 0β (
g jβ,i + giβ, j − gi j ,β)
= 1
2g00,t + 1
2hi j gi j ,t
where gi j ,t = 2ki j in view of the above. Thus, we define
g00,t :=−2trhk.
From the wave coordinate conditions gαβΓlαβ
= 0, we would have on P
gαβΓlαβ = −Γl
00 +hi jΓli j
= −1
2g lα (
gα0,0 + g0α,0 − g00,α)+hi j 1
2g lβ (
g jβ,i + giβ, j − gi j ,β)
= −1
2g lm (
gm0,0 + g0m,0 − g00,m)+hi j 1
2g lm (
g j m,i + gi m, j − gi j ,m)
= −hl m gm0,0 +hi j 1
2hlm (
g j m,i + gi m, j − gi j ,m)
.
Thus, we need
gm0,0 := hi j 1
2
(g j m,i + gi m, j − gi j ,m
).
Since we now have enough data, we can apply a local existence theorem forquasilinear wave to obtain the existence of a metric g , solution of the reduced equa-tions on some set M of the required form. However, so far the metric g only satisfiesthe reduced Einstein equations and we need to show that the wave coordinate con-ditions is indeed satisfied by such a g . It then would follows that g actually solvesthe Einstein equations.
Recall from the section on the wave coordinate conditions that the wave condi-tions can be equivalently formulated as
Γα := gµνΓµαν = 0.
Going back to the computation of the Ricci tensor in wave coordinates, one hasthat, in general,
Ri c(g )µν =−1
2gαβ∂αβgµν+Qµν(∂g ,∂g )+ 1
2
(DµΓν+DνΓµ
).
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From this, it follows that the Einstein tensor verifies
Gµν = −1
2gαβ∂αβgµν+Qµν(∂g ,∂g )+ 1
2
(∂µΓν+∂νΓµ
)− 1
2(∂αΓα)gµν
= 1
2
(∂µΓν+∂νΓµ
)− 1
2(∂αΓα)gµν,
since g solves the reduced Einstein equations.Recall now that the Einstein tensor is always divergence free and thus that
DµGµν = 0.
In view of the above, we obtain a system of wave equations of the form
gαβ∂αβΓν = F (∂Γν).
In order for Γν to vanish, it follows that it suffices to show that initially Γν = 0 and∂tΓν = 0.
That Γν = 0 on P is certainly true since we have set up our initial data for this tohold. For the derivative conditions, since P is a submanifold of M and g a Lorentzianmetric, the Gauss-Codazzi equations (24)-(25) must hold for the metric g on P i.e.
G(N , N ) = 1
2SP −ki j k i j + (trhk)2
G(N , v) = DP j kki − DPi (trhk)v i
On the other hand, the RHS vanishes by virtue of the constraint equations (35)-(35).Thus, we have G(N , N ) = 0 and G(N , v) = 0, where N = ∂t . The first condition gives∂tΓ0 = 0. The second, gives with v = ∂xi ,
G0i = 1
2(∂tΓi +∂iΓ0)
= 1
2∂tΓi .
Thus, we have proven that the Γν all have trivial data and in view of the above discus-sion, it follows that g solves the vacuum Einstein equations in set M of the requiredform.
A Global volume form and orientable manifolds
Lemma A.1. A semi-Riemannian manifold M has a global volume form if and onlyif M is orientable.
Proof. Recall that a manifold is orientable if it has a coordinate atlas all of whosetransition functions have positive Jacobian determinants. For every such local coor-dinate system, we have a local n-form defined by |det(g )|d x1∧..∧d xn . Moreover, forevery other coordinate in the atlas (thus positively oriented with the first) one cancheck that the local n-form in the new coordinate system agrees on overlap with thefirst. Thus, this defines a global volume form.
120
B The pull-back bundle
Recall first the definition of a smooth fiber bundle. Let E ,B ,F be smooth manifoldsand π : E → B be a smooth surjective map such that for all x ∈ E , there is an openneighbourhood U ⊂ B of π(x) together with a diffeomorphism φ : π−1(U ) → U ×Fsatisfying π=πU φ, where πU : U ×F →U is the projection on the first factor.
Let π : E → M be a fiber bundle over M and consider a smooth map
φ : N → M .
We define the pull-back bundle as
φ?(E) = (q,e) ∈ N ×E |φ(q) =π(e)
⊂ N ×E ,
equipped with the subspace topology and the projection map
π′ :φ?E → N
(q,e) → q.
C Not so useful facts
We state here too standard results concerning the existence of a Lorentzian metric.
Theorem C.1. A compact manifold admits a Lorentzian metric if and only if its Eulercaracteristic is 0.
Theorem C.2. Every non-compact manifold admits a Lorentzian metric.
The proof of this theorem is not difficult using the existence of a Riemannianmetric on M (which is true for all manifolds) together with the following statement.
Theorem C.3. Every non-compact manifold admits a real-valued function f whosedifferential is nowhere vanishing.
Note that in General Relativity, Lorentzian manifold are constructed by solvingthe Einstein equations (in particular, the underlying manifold is never compact!), sothat the above statements are seldom (never ? ) used.
D Taylor expansions of the metric tensor and volumeform, following Alfred Gray [Gra73, Gra04]
We consider a semi-Riemannian manifold (M , g ). Given vector fields X , X1, .., Xq wewrite
DqX1 X2...Xq
X := DX1 DX2 ..DXq X .
Let (xα) denotes normal coordinates at some point p ∈ M . We consider normalcoordinate vector fields, i.e. vector fields whose components in the ∂xα basis areconstant. From Exercise 3.17, the integral curve of any such vector field through p isa geodesic.
We will need the following lemmas.
121
Lemma D.1. Let X ,Y be a normal coordinate vector fields and ξ the integral curve ofX through p. Then, (
DqX ..X ..X X
)ξ(t ) = 0,
(DX Y )p = 0.
Proof. We have (DX X )ξ(t ) = 0, since ξ(t ) is a geodesic and ξ = X ξ. Moreover, bydefinition of the induced covariant derivative, we have
0 = d
d s
[(DX X )ξ(t )
]= D ξDX X = D2X X X .
By induction, we obtain the first statement of the lemma. Now, since (DX X )p = 0,we have in particular,
0 = (DX+Y (X +Y ))p = (DX Y )p + (DY X )p + (DY Y )p + (DX X )p
= (DX Y )p + (DY X )p .
On the other hand, since X ,Y have constant components with respect to a coordi-nate induced basis of vector fields, they commute, and the second statement of thelemma follows.
Lemma D.2. Let X ,Y be normal coordinate vector fields. Then,
(Dq
Y X ...X X)
p = .. = (Dp
X ..X Y X X)
p ,(2Dq
X ..X Y + (q −1)DqX X X Y X X
)p = 0,(
DqX ..X Y
)p = q −1
q +1Dq−2
X ..X R(X ,Y )X .
Proof. Since [X ,Y ] = 0, we have R(X ,Y ) = DX DY −DY DX . Let Ak := (Dq
X ..Y ..X X)
p ,
where the Y is in k t h place.We have, for k ≥ 2
Ak − Ak−1 = Dk−2X ..X R(X ,Y )Dq−k
X ..X X .
Note that since R is a tensor, we have, if q > k(R(X ,Y )Dq−k
X ..X X)ξ(t )
= 0,
since(Dq−k
X ..X X)ξ(t )
= 0. It then follows by induction that Dk−2X ..X R(X ,Y )Dq−k
X ..X X = 0, for
any q > k, which is the first statement of the lemma.For the second statement, we have by the first lemma, that for any t ∈R,(
DqX+tY (X + tY )
)p = 0.
The left hand side of the previous equation defined polynom in t and we computeits linear coefficient.(Dq
(X+tY )..(X+tY )(X + tY ))
p= Dq
(X+tY )..(X+tY )X + tDq(X+tY )..(X+tY )Y
= Dq−1(X+tY )..(X+tY )DX X + tDq−1
(X+tY )..(X+tY )DY X + tDpX ..X Y +Dq
Y ..Y Y +O(t 2)
= t (q −1)Dq−1Y X ..X DX X + tDq−1
X ..X DY X + tDpX ..X Y +O(t 2)
= t (q −1)Dq−1Y X ..X DX X +2tDp
X ..X Y +O(t 2),
122
where all terms should be evaluated at p and we have used that DqY ..Y Y = 0, the
first statement of the lemma, and the fact that DX Y = DY X . This proves the secondstatement of the lemma.
For the last statement, we start from the second, which we write as(2Dq
X ..X Y)
p =−(q −1)(Dq−2
X X X DY DX X)
p.
We add (q −1)(Dq−2
X X X DX DY X)
pto both side of the equations, so to introduce the
Riemann curvature tensor on the right-hand side(2Dq
X ..X Y)
p + (q −1)(Dq−2
X X X DX DY X)
p= (q −1)
(Dq−2
X X X R(X ,Y )X)
p.
The formula then follows using DX Y = DY X .
Lemma D.3. Let X1, X2 and Y be normal coordinate vector fields. Then, we have
DX1 DX2 Y +DX1 DX2 Y = 1
3(R(X1,Y )X2 +R(X2,Y )Y ) .
Proof. Apply the third statement of the previous lemma to X1+t X2 and Y with q = 2and then compute the coefficient linear in t .
Consider now the local volume form η
η=√
|det g |d x1 ∧ ...∧d xn .
We recall that Dη = 0. We consider a taylor expansion near p = (xα = 0) forp
g =η
(∂x1 , ...,∂xn
).
η(∂x1 , ...,∂xn
)(xα) = η
(∂x1 , ...,∂xn
)(0)+∂xα
(η
(∂x1 , ...,∂xn
))(0)xα
+ 1
2∂xα∂xβ
(η
(∂x1 , ...,∂xn
))(0)xαxβ+O(|x|3).
We have η(∂x1 , ...,∂xn
)(0) = 1 while
∂xα(η
(∂x1 , ...,∂xn
))(0) = (
(Dαη)(∂x1 , ...,∂xn
))(0)+η(∂x1 , ..,Dα∂xβ , ..,∂xn )(0) = 0
since Dη= 0 and Dα∂xβ (0) = 0. For the second order term, we compute similarly
Recall now that we have, by definition of the components of R
R(∂xα ,∂xγ )∂xβ = Rρ
βαγ∂xρ .
Moreover, by antisymmetry of η, if the curvature terms on the RHS of (35) are in γthposition, only the term proportional to ∂xγ will contribute. Thus, we have
1
2∂xα∂xβ
(η
(∂x1 , ...,∂xn
))(0)xαxβ = 1
12
(Rγ
βαγ+Rγ
αβγ
)(0)xαxβ
= −1
6Ri c(g )αβ(0)xαxβ.
We have thus proven the following theorem.
Theorem D.1. With respect to a normal coordinate system (xα) at p, the volume formadmits the taylor expansion
η(x) =(1− 1
6Ri c(g )αβ(0)xαxβ+O(|x|3)
)d x1 ∧ ...∧d xn .
E Additional exercices
E.1 Problems
E.1.1 A commutation formula for the wave equation
Let äg denote the D’Alembertian acting on real valued functions ψ : M →R.
1. Let X be a Killing field. Prove that [X ,äg ] = 0, in the sense that for any smoothfunction ψ,
X (ägψ) =äg (X (ψ)).
2. Let now X be a vector field, not necessarily Killing and denote by π its defor-mation tensor, and by tr (π) the trace ofπ, given in any local coordinate systemby tr (π) = gαβπαβ. Prove that for any smooth function ψ;
g(Xψ
)= X (äg (ψ))+q[
Xψ]
with
q[
Xψ]= 2παβDαDβψ+ [
2Dαπαµ−Dµ (trπ)]
Dµψ .
E.2 Solutions
E.2.1 Solution to E.1.1 (second question only)
This is a classical formula. The proof written below is taken from [HS13]), Lemma6.2, p242.
Note that on functions f
LX Dα f = DαLX = X βDβDα f +(DαX β
)Dβ f ,
while on 1-forms
LX DαVβ−DαLX Vβ = X γ[Dγ,Dα]Vβ−V γ[Dγ,Dα]Xβ+(DγDαXβ−2Dαπγβ
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